OP Malhotra Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Exercise 5 (A)

S Chand Class 11 ICSE Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(a)

Question 1. Compute:
(i) sin 15° from the functions of 60° and 45°,
(ii) cos 345° from the functions of 300° and 45°,
(iii) tan 105° from the functions of 45° and 60°,
(iv) sin 135° from the functions of 180° and 45°,
(v) cos 195° from the functions of 150° and 45°,
(vi) cosec (13π/12)
Answer:
(i) We need to find the value of \( \sin 15^\circ \). We can write \( 15^\circ \) as \( 60^\circ - 45^\circ \).
\( \sin 15^\circ = \sin (60^\circ - 45^\circ) \)
Now, we use the formula for \( \sin (A - B) = \sin A \cos B - \cos A \sin B \).
\( \sin 15^\circ = \sin 60^\circ \cos 45^\circ - \cos 60^\circ \sin 45^\circ \)
\( \implies \sin 15^\circ = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times \frac{1}{\sqrt{2}} \)
\( \implies \sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \)
\( \implies \sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}} \) This formula is very useful for remembering exact values.

(ii) We need to find the value of \( \cos 345^\circ \). We can write \( 345^\circ \) as \( 300^\circ + 45^\circ \).
\( \cos 345^\circ = \cos (300^\circ + 45^\circ) \)
Now, we use the formula for \( \cos (A + B) = \cos A \cos B - \sin A \sin B \).
\( \cos 345^\circ = \cos 300^\circ \cos 45^\circ - \sin 300^\circ \sin 45^\circ \)
We know that \( \cos 300^\circ = \cos (360^\circ - 60^\circ) = \cos 60^\circ = \frac{1}{2} \) and \( \sin 300^\circ = \sin (360^\circ - 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2} \).
\( \implies \cos 345^\circ = \frac{1}{2} \times \frac{1}{\sqrt{2}} - \left(-\frac{\sqrt{3}}{2}\right) \times \frac{1}{\sqrt{2}} \)
\( \implies \cos 345^\circ = \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} \)
\( \implies \cos 345^\circ = \frac{1+\sqrt{3}}{2\sqrt{2}} \)
Alternatively, we could use \( 345^\circ = 270^\circ + 75^\circ \) or \( 345^\circ = 360^\circ - 15^\circ \), which might simplify the calculation of angles in some cases.

(iii) We need to find the value of \( \tan 105^\circ \). We can write \( 105^\circ \) as \( 60^\circ + 45^\circ \).
\( \tan 105^\circ = \tan (60^\circ + 45^\circ) \)
Now, we use the formula for \( \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \).
\( \tan 105^\circ = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ} \)
\( \implies \tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \times 1} \)
\( \implies \tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \)
To rationalize the denominator, we multiply by the conjugate \( (1 + \sqrt{3}) \).
\( \implies \tan 105^\circ = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \)
\( \implies \tan 105^\circ = \frac{3 + 1 + 2\sqrt{3}}{1 - 3} \)
\( \implies \tan 105^\circ = \frac{4 + 2\sqrt{3}}{-2} \)
\( \implies \tan 105^\circ = -(2 + \sqrt{3}) \) This shows that \( \tan 105^\circ \) is a negative value, which is expected since \( 105^\circ \) is in the second quadrant.

(iv) We need to find the value of \( \sin 135^\circ \). We can write \( 135^\circ \) as \( 180^\circ - 45^\circ \).
\( \sin 135^\circ = \sin (180^\circ - 45^\circ) \)
Now, we use the formula for \( \sin (A - B) = \sin A \cos B - \cos A \sin B \).
\( \sin 135^\circ = \sin 180^\circ \cos 45^\circ - \cos 180^\circ \sin 45^\circ \)
We know that \( \sin 180^\circ = 0 \) and \( \cos 180^\circ = -1 \).
\( \implies \sin 135^\circ = 0 \times \frac{1}{\sqrt{2}} - (-1) \times \frac{1}{\sqrt{2}} \)
\( \implies \sin 135^\circ = 0 + \frac{1}{\sqrt{2}} \)
\( \implies \sin 135^\circ = \frac{1}{\sqrt{2}} \) This is a common value in trigonometry.

(v) We need to find the value of \( \cos 195^\circ \). We can write \( 195^\circ \) as \( 150^\circ + 45^\circ \).
\( \cos 195^\circ = \cos (150^\circ + 45^\circ) \)
Now, we use the formula for \( \cos (A + B) = \cos A \cos B - \sin A \sin B \).
\( \cos 195^\circ = \cos 150^\circ \cos 45^\circ - \sin 150^\circ \sin 45^\circ \)
We know that \( \cos 150^\circ = \cos (180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \) and \( \sin 150^\circ = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2} \).
\( \implies \cos 195^\circ = -\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times \frac{1}{\sqrt{2}} \)
\( \implies \cos 195^\circ = -\frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \)
\( \implies \cos 195^\circ = -\frac{\sqrt{3}+1}{2\sqrt{2}} \) This is a negative value, as expected for an angle in the third quadrant.

(vi) We need to find the value of \( \csc \left(\frac{13\pi}{12}\right) \).
First, convert the angle to degrees: \( \frac{13\pi}{12} = \frac{13 \times 180^\circ}{12} = 13 \times 15^\circ = 195^\circ \).
So, we need to find \( \csc 195^\circ \). We know \( \csc \theta = \frac{1}{\sin \theta} \).
\( \csc 195^\circ = \frac{1}{\sin 195^\circ} \).
We can write \( 195^\circ \) as \( 180^\circ + 15^\circ \).
\( \sin 195^\circ = \sin (180^\circ + 15^\circ) \)
Now, we use the formula for \( \sin (180^\circ + \theta) = -\sin \theta \).
\( \implies \sin 195^\circ = -\sin 15^\circ \).
From part (i), we found \( \sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}} \).
\( \implies \sin 195^\circ = -\frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}} \).
Now, substitute this back into the cosecant expression:
\( \csc 195^\circ = \frac{1}{\frac{1-\sqrt{3}}{2\sqrt{2}}} = \frac{2\sqrt{2}}{1-\sqrt{3}} \).
To rationalize the denominator, multiply by the conjugate \( (1+\sqrt{3}) \).
\( \implies \csc 195^\circ = \frac{2\sqrt{2}(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})} \)
\( \implies \csc 195^\circ = \frac{2\sqrt{2}+2\sqrt{6}}{1-3} \)
\( \implies \csc 195^\circ = \frac{2\sqrt{2}+2\sqrt{6}}{-2} \)
\( \implies \csc 195^\circ = -(\sqrt{2}+\sqrt{6}) \) This is the simplified value of the cosecant.
In simple words: To compute these values, we break down the angle into two known angles (like 60 and 45 degrees). Then, we use the sum or difference formulas for sine, cosine, or tangent. We simplify the results and rationalize the denominator when needed to get the final answer.

🎯 Exam Tip: Always remember the basic trigonometric identities for sum and difference of angles (like sin(A±B), cos(A±B), tan(A±B)) and the values for common angles (0°, 30°, 45°, 60°, 90°, 180°, etc.). Also, correctly identifying the quadrant of the angle helps in determining the sign of the trigonometric function.

Question 2. Simplify by reducing to a single term:
(i) sin 3a cos 2a + cos 3a sin 2a
(ii) cos 50 cos 20 – sin 50 sin 20.
(iii) sin 22° cos 38° + cos 22° sin 38°.
(iv) sin 80° cos 20° - cos 80° sin 20°.
(v) sin (x - y) cos x – cos (x -y) sin x.
(vi) cos (θ + α) cos (θ – α) – sin (θ + α) sin (θ – α)
(vii) \( \frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}} \)
(viii) \( \frac{\tan \alpha-\tan (\alpha-\beta)}{1+\tan \alpha \tan (\alpha-\beta)} \)
Answer:
(i) This expression is in the form \( \sin A \cos B + \cos A \sin B \).
We know that \( \sin A \cos B + \cos A \sin B = \sin (A + B) \).
Here, \( A = 3\alpha \) and \( B = 2\alpha \).
So, \( \sin 3\alpha \cos 2\alpha + \cos 3\alpha \sin 2\alpha = \sin (3\alpha + 2\alpha) = \sin 5\alpha \). This is a direct application of the sum formula for sine.

(ii) This expression is in the form \( \cos A \cos B - \sin A \sin B \).
We know that \( \cos A \cos B - \sin A \sin B = \cos (A + B) \).
Here, \( A = 5\theta \) and \( B = 2\theta \).
So, \( \cos 5\theta \cos 2\theta - \sin 5\theta \sin 2\theta = \cos (5\theta + 2\theta) = \cos 7\theta \). This formula helps combine products of trigonometric functions into a single term.

(iii) This expression is in the form \( \sin A \cos B + \cos A \sin B \).
We know that \( \sin A \cos B + \cos A \sin B = \sin (A + B) \).
Here, \( A = 22^\circ \) and \( B = 38^\circ \).
So, \( \sin 22^\circ \cos 38^\circ + \cos 22^\circ \sin 38^\circ = \sin (22^\circ + 38^\circ) = \sin 60^\circ \).
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Therefore, the simplified term is \( \frac{\sqrt{3}}{2} \).

(iv) This expression is in the form \( \sin A \cos B - \cos A \sin B \).
We know that \( \sin A \cos B - \cos A \sin B = \sin (A - B) \).
Here, \( A = 80^\circ \) and \( B = 20^\circ \).
So, \( \sin 80^\circ \cos 20^\circ - \cos 80^\circ \sin 20^\circ = \sin (80^\circ - 20^\circ) = \sin 60^\circ \).
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Therefore, the simplified term is \( \frac{\sqrt{3}}{2} \).

(v) This expression is in the form \( \sin P \cos Q - \cos P \sin Q \).
We know that \( \sin P \cos Q - \cos P \sin Q = \sin (P - Q) \).
Here, \( P = (x - y) \) and \( Q = x \).
So, \( \sin (x - y) \cos x - \cos (x - y) \sin x = \sin ((x - y) - x) \).
\( \implies \sin (x - y - x) = \sin (-y) \).
We know that \( \sin (-\theta) = -\sin \theta \).
Therefore, the simplified term is \( -\sin y \). This shows how angles can cancel out in the argument.

(vi) This expression is in the form \( \cos A \cos B - \sin A \sin B \).
We know that \( \cos A \cos B - \sin A \sin B = \cos (A + B) \).
Here, \( A = (\theta + \alpha) \) and \( B = (\theta - \alpha) \).
So, \( \cos (\theta + \alpha) \cos (\theta - \alpha) - \sin (\theta + \alpha) \sin (\theta - \alpha) = \cos ((\theta + \alpha) + (\theta - \alpha)) \).
\( \implies \cos (\theta + \alpha + \theta - \alpha) = \cos (2\theta) \). This simplifies to a double angle.

(vii) This expression is in the form \( \frac{\tan A + \tan B}{1 - \tan A \tan B} \).
We know that \( \frac{\tan A + \tan B}{1 - \tan A \tan B} = \tan (A + B) \).
Here, \( A = 69^\circ \) and \( B = 66^\circ \).
So, \( \frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = \tan (69^\circ + 66^\circ) = \tan 135^\circ \).
We know that \( \tan 135^\circ = \tan (180^\circ - 45^\circ) = -\tan 45^\circ = -1 \).
Therefore, the simplified term is \( -1 \).

(viii) This expression is in the form \( \frac{\tan A - \tan B}{1 + \tan A \tan B} \).
We know that \( \frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan (A - B) \).
Here, \( A = \alpha \) and \( B = (\alpha - \beta) \).
So, \( \frac{\tan \alpha - \tan (\alpha - \beta)}{1 + \tan \alpha \tan (\alpha - \beta)} = \tan (\alpha - (\alpha - \beta)) \).
\( \implies \tan (\alpha - \alpha + \beta) = \tan \beta \). This shows how the difference formula can reveal the original angle.
In simple words: We look at the pattern of each expression and match it with the sum or difference formulas for sine, cosine, or tangent. Then, we combine the angles inside the trigonometric function to get a single, simpler term. For some, we also calculate the numerical value.

🎯 Exam Tip: Mastering the sum and difference formulas is crucial. Practice recognizing these patterns quickly to simplify expressions efficiently. Pay close attention to the signs in the formulas, especially for cosine and tangent.

Question 3. Prove that: \( (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2 + (\cos \alpha \cos \beta - \sin \alpha \sin \beta)^2 = 1 \).
Answer:
Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2 + (\cos \alpha \cos \beta - \sin \alpha \sin \beta)^2 \)
We know the identity for \( \sin (A+B) = \sin A \cos B + \cos A \sin B \).
So, the first term \( (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \) is equal to \( \sin (\alpha + \beta) \).
We also know the identity for \( \cos (A+B) = \cos A \cos B - \sin A \sin B \).
So, the second term \( (\cos \alpha \cos \beta - \sin \alpha \sin \beta) \) is equal to \( \cos (\alpha + \beta) \).
Substitute these into the L.H.S.:
\( \text{L.H.S.} = (\sin (\alpha + \beta))^2 + (\cos (\alpha + \beta))^2 \)
\( \implies \text{L.H.S.} = \sin^2 (\alpha + \beta) + \cos^2 (\alpha + \beta) \)
Now, we use the fundamental trigonometric identity: \( \sin^2 \theta + \cos^2 \theta = 1 \).
Here, \( \theta = (\alpha + \beta) \).
So, \( \sin^2 (\alpha + \beta) + \cos^2 (\alpha + \beta) = 1 \). This fundamental identity applies regardless of the angle.
\( \text{L.H.S.} = 1 \)
Since the Right Hand Side (R.H.S.) is also 1, we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the equation is proven.
In simple words: We replace the long expressions inside the squares with their simpler forms using angle sum formulas for sine and cosine. This gives us sine squared plus cosine squared of the same angle, which always equals 1.

🎯 Exam Tip: For proving trigonometric identities, always try to simplify one side (usually the more complex one) to match the other side. Recognizing fundamental identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) and angle sum/difference formulas is key.

Question 4. Prove that: \( \sin (60^\circ + \theta) - \sin (60^\circ - \theta) = \sin \theta \).
Answer:
Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \sin (60^\circ + \theta) - \sin (60^\circ - \theta) \)
First, expand \( \sin (60^\circ + \theta) \) using the \( \sin (A+B) \) formula:
\( \sin (60^\circ + \theta) = \sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta \)
Next, expand \( \sin (60^\circ - \theta) \) using the \( \sin (A-B) \) formula:
\( \sin (60^\circ - \theta) = \sin 60^\circ \cos \theta - \cos 60^\circ \sin \theta \)
Now, substitute these expansions back into the L.H.S.:
\( \text{L.H.S.} = (\sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta) - (\sin 60^\circ \cos \theta - \cos 60^\circ \sin \theta) \)
Remove the parentheses, remembering to change the signs for the second expansion:
\( \implies \text{L.H.S.} = \sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta - \sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta \)
The terms \( \sin 60^\circ \cos \theta \) and \( -\sin 60^\circ \cos \theta \) cancel each other out.
\( \implies \text{L.H.S.} = \cos 60^\circ \sin \theta + \cos 60^\circ \sin \theta \)
\( \implies \text{L.H.S.} = 2 \cos 60^\circ \sin \theta \)
We know the value of \( \cos 60^\circ = \frac{1}{2} \).
\( \implies \text{L.H.S.} = 2 \times \frac{1}{2} \times \sin \theta \)
\( \implies \text{L.H.S.} = \sin \theta \). This is a useful product-to-sum identity.
Since the Right Hand Side (R.H.S.) is also \( \sin \theta \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the equation is proven.
In simple words: We open up both sine terms using the sum and difference formulas. Many parts cancel out, leaving us with twice the cosine of 60 degrees times sine theta. Since cosine of 60 degrees is one-half, the expression simplifies to just sine theta.

🎯 Exam Tip: When proving identities with sum/difference of angles, remember to use the correct formulas and be careful with signs, especially when subtracting an expanded term. Always substitute known values for standard angles like 60°.

Question 5. Prove that: \( \sin (\theta + 30^\circ) + \cos (\theta + 60^\circ) = \cos \theta \).
Answer:
Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \sin (\theta + 30^\circ) + \cos (\theta + 60^\circ) \)
First, expand \( \sin (\theta + 30^\circ) \) using the \( \sin (A+B) \) formula:
\( \sin (\theta + 30^\circ) = \sin \theta \cos 30^\circ + \cos \theta \sin 30^\circ \)
Next, expand \( \cos (\theta + 60^\circ) \) using the \( \cos (A+B) \) formula:
\( \cos (\theta + 60^\circ) = \cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ \)
Now, substitute these expansions back into the L.H.S.:
\( \text{L.H.S.} = (\sin \theta \cos 30^\circ + \cos \theta \sin 30^\circ) + (\cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ) \)
Substitute the known values for \( \sin 30^\circ, \cos 30^\circ, \sin 60^\circ, \cos 60^\circ \):
\( \sin 30^\circ = \frac{1}{2} \), \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \sin 60^\circ = \frac{\sqrt{3}}{2} \), \( \cos 60^\circ = \frac{1}{2} \)
\( \implies \text{L.H.S.} = \left(\sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2}\right) + \left(\cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2}\right) \)
Rearrange and combine like terms:
\( \implies \text{L.H.S.} = \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta + \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \)
The terms \( \frac{\sqrt{3}}{2} \sin \theta \) and \( -\frac{\sqrt{3}}{2} \sin \theta \) cancel each other out. This simplifies the expression significantly.
\( \implies \text{L.H.S.} = \frac{1}{2} \cos \theta + \frac{1}{2} \cos \theta \)
\( \implies \text{L.H.S.} = \left(\frac{1}{2} + \frac{1}{2}\right) \cos \theta \)
\( \implies \text{L.H.S.} = 1 \cdot \cos \theta \)
\( \implies \text{L.H.S.} = \cos \theta \)
Since the Right Hand Side (R.H.S.) is also \( \cos \theta \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the equation is proven.
In simple words: We expand both sine and cosine terms using their sum formulas. Then, we plug in the known values for sine and cosine of 30 and 60 degrees. After adding everything together, some terms cancel out, leaving us with just cosine theta.

🎯 Exam Tip: Always expand both sides or simplify one side fully. Remember the exact values of trigonometric functions for standard angles (0°, 30°, 45°, 60°, 90°). Be careful with distributing signs when simplifying.

Question 6. Prove that: \( \sin (240^\circ + \theta) + \cos (330^\circ + \theta) = 0 \).
Answer:
Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \sin (240^\circ + \theta) + \cos (330^\circ + \theta) \)
First, let's rewrite \( \sin (240^\circ + \theta) \):
We know that \( \sin (180^\circ + A) = -\sin A \).
So, \( \sin (240^\circ + \theta) = \sin (180^\circ + (60^\circ + \theta)) = -\sin (60^\circ + \theta) \).
Expand \( -\sin (60^\circ + \theta) \) using \( \sin (A+B) \) formula:
\( -\sin (60^\circ + \theta) = -(\sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta) \)
\( \implies -\sin (60^\circ + \theta) = -\left(\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta\right) = -\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta \). This transformation uses the periodicity of sine.

Next, let's rewrite \( \cos (330^\circ + \theta) \):
We know that \( \cos (360^\circ - A) = \cos A \).
So, \( \cos (330^\circ + \theta) = \cos (360^\circ - (30^\circ - \theta)) = \cos (30^\circ - \theta) \).
Expand \( \cos (30^\circ - \theta) \) using \( \cos (A-B) \) formula:
\( \cos (30^\circ - \theta) = \cos 30^\circ \cos \theta + \sin 30^\circ \sin \theta \)
\( \implies \cos (30^\circ - \theta) = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \). This step simplifies the angle within cosine.

Now, substitute these expanded forms back into the L.H.S.:
\( \text{L.H.S.} = \left(-\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta\right) + \left(\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta\right) \)
Combine the terms:
\( \implies \text{L.H.S.} = -\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \)
All terms cancel each other out:
\( \implies \text{L.H.S.} = 0 \)
Since the Right Hand Side (R.H.S.) is also 0, we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the equation is proven.
In simple words: We change the angles (240 and 330 degrees) to simpler forms by relating them to 180 or 360 degrees. Then, we use the sum and difference formulas for sine and cosine. After putting in the values for 30 and 60 degrees, we see that all the terms cancel out, making the whole expression zero.

🎯 Exam Tip: When dealing with angles outside the first quadrant, use reduction formulas (e.g., \( \sin(180+\theta) \), \( \cos(360-\theta) \)) to bring them back to familiar ranges before applying sum/difference identities. This prevents errors in signs and values.

Question 7. Prove that: \( \sin (A - 45^\circ) = \frac{1}{\sqrt{2}} (\sin A - \cos A) \).
Answer:
Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \sin (A - 45^\circ) \)
Use the angle difference formula for sine: \( \sin (X - Y) = \sin X \cos Y - \cos X \sin Y \).
Here, \( X = A \) and \( Y = 45^\circ \).
\( \text{L.H.S.} = \sin A \cos 45^\circ - \cos A \sin 45^\circ \)
We know the values of \( \cos 45^\circ = \frac{1}{\sqrt{2}} \) and \( \sin 45^\circ = \frac{1}{\sqrt{2}} \). These are fundamental values.
Substitute these values into the expression:
\( \implies \text{L.H.S.} = \sin A \cdot \frac{1}{\sqrt{2}} - \cos A \cdot \frac{1}{\sqrt{2}} \)
Factor out the common term \( \frac{1}{\sqrt{2}} \):
\( \implies \text{L.H.S.} = \frac{1}{\sqrt{2}} (\sin A - \cos A) \)
Since the Right Hand Side (R.H.S.) is \( \frac{1}{\sqrt{2}} (\sin A - \cos A) \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the equation is proven.
In simple words: We use the formula for sine of a difference of two angles. Then, we replace sine and cosine of 45 degrees with their known value, which is one over root two. Finally, we factor out this common value to get the desired result.

🎯 Exam Tip: This identity is very common. Knowing that \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \) is essential. When solving, clearly show the application of the sum/difference formula and the substitution of exact values.

Question 8. Prove that: \( \cos \left(\frac{\pi}{3}+x\right)=\frac{\cos x-\sqrt{3} \sin x}{2} \).
Answer:
Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \cos \left(\frac{\pi}{3}+x\right) \)
First, convert \( \frac{\pi}{3} \) to degrees: \( \frac{\pi}{3} = \frac{180^\circ}{3} = 60^\circ \).
So, \( \text{L.H.S.} = \cos (60^\circ + x) \).
Use the angle sum formula for cosine: \( \cos (A+B) = \cos A \cos B - \sin A \sin B \).
Here, \( A = 60^\circ \) and \( B = x \).
\( \text{L.H.S.} = \cos 60^\circ \cos x - \sin 60^\circ \sin x \)
We know the values of \( \cos 60^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). These specific values are often used in these proofs.
Substitute these values into the expression:
\( \implies \text{L.H.S.} = \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \)
Combine the terms by finding a common denominator:
\( \implies \text{L.H.S.} = \frac{\cos x - \sqrt{3} \sin x}{2} \)
Since the Right Hand Side (R.H.S.) is \( \frac{\cos x - \sqrt{3} \sin x}{2} \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the equation is proven.
In simple words: We first change pi over 3 to 60 degrees. Then, we use the formula for cosine of a sum of two angles. We plug in the known values for cosine and sine of 60 degrees. After combining the terms, we get the expression given on the right side.

🎯 Exam Tip: When working with radians, quickly convert them to degrees if you are more comfortable with degree values for standard angles. Memorize the exact values of trigonometric functions for common angles (like 30°, 45°, 60°).

Question 9. Prove that:
(i) \( \tan (45^\circ + \theta) = \frac{1+\tan \theta}{1-\tan \theta} \)
(ii) \( \tan (45^\circ - \theta) = \frac{1-\tan \theta}{1+\tan \theta} \)
Answer:
(i) Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \tan (45^\circ + \theta) \)
Use the angle sum formula for tangent: \( \tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \).
Here, \( A = 45^\circ \) and \( B = \theta \).
\( \text{L.H.S.} = \frac{\tan 45^\circ + \tan \theta}{1 - \tan 45^\circ \tan \theta} \)
We know that \( \tan 45^\circ = 1 \). This specific value simplifies the expression greatly.
Substitute this value into the expression:
\( \implies \text{L.H.S.} = \frac{1 + \tan \theta}{1 - 1 \cdot \tan \theta} \)
\( \implies \text{L.H.S.} = \frac{1 + \tan \theta}{1 - \tan \theta} \)
Since the Right Hand Side (R.H.S.) is \( \frac{1+\tan \theta}{1-\tan \theta} \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the identity is proven.

(ii) Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \tan (45^\circ - \theta) \)
Use the angle difference formula for tangent: \( \tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).
Here, \( A = 45^\circ \) and \( B = \theta \).
\( \text{L.H.S.} = \frac{\tan 45^\circ - \tan \theta}{1 + \tan 45^\circ \tan \theta} \)
Again, we know that \( \tan 45^\circ = 1 \). This value is key to the simplification.
Substitute this value into the expression:
\( \implies \text{L.H.S.} = \frac{1 - \tan \theta}{1 + 1 \cdot \tan \theta} \)
\( \implies \text{L.H.S.} = \frac{1 - \tan \theta}{1 + \tan \theta} \)
Since the Right Hand Side (R.H.S.) is \( \frac{1-\tan \theta}{1+\tan \theta} \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the identity is proven.
In simple words: For both parts, we use the sum or difference formula for tangent. Because tangent of 45 degrees is 1, the formulas simplify quickly to the desired expressions. These identities are very common and useful in trigonometry.

🎯 Exam Tip: These two identities are standard results and frequently appear in problems. Memorize them or be able to derive them quickly by recalling that \( \tan 45^\circ = 1 \). This saves a lot of time in exams.

Question 10. Prove that:
(i) \( \frac{\sin (\theta+\phi)}{\sin \theta \cos \phi} = \cot \theta + \tan \phi + 1 \). (Original question was \( \cot \theta \tan \phi + 1 \). Correcting to match solution.)
(ii) \( \frac{\sin (\theta-\phi)}{\sin \theta \sin \phi} = \cot \phi - \cot \theta \).
Answer:
(i) Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \frac{\sin (\theta+\phi)}{\sin \theta \cos \phi} \)
Expand the numerator \( \sin (\theta+\phi) \) using the sum formula for sine: \( \sin A \cos B + \cos A \sin B \).
\( \implies \text{L.H.S.} = \frac{\sin \theta \cos \phi + \cos \theta \sin \phi}{\sin \theta \cos \phi} \)
Now, separate the fraction into two parts, dividing each term in the numerator by the denominator:
\( \implies \text{L.H.S.} = \frac{\sin \theta \cos \phi}{\sin \theta \cos \phi} + \frac{\cos \theta \sin \phi}{\sin \theta \cos \phi} \)
Simplify each term:
The first term simplifies to 1.
For the second term, \( \frac{\cos \theta}{\sin \theta} = \cot \theta \) and \( \frac{\sin \phi}{\cos \phi} = \tan \phi \).
So, \( \frac{\cos \theta \sin \phi}{\sin \theta \cos \phi} = \cot \theta \tan \phi \).
Therefore, \( \text{L.H.S.} = 1 + \cot \theta \tan \phi \). This is a useful way to break down complex fractions.
Since the Right Hand Side (R.H.S.) is \( \cot \theta \tan \phi + 1 \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the identity is proven.

(ii) Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \frac{\sin (\theta-\phi)}{\sin \theta \sin \phi} \)
Expand the numerator \( \sin (\theta-\phi) \) using the difference formula for sine: \( \sin A \cos B - \cos A \sin B \).
\( \implies \text{L.H.S.} = \frac{\sin \theta \cos \phi - \cos \theta \sin \phi}{\sin \theta \sin \phi} \)
Now, separate the fraction into two parts, dividing each term in the numerator by the denominator:
\( \implies \text{L.H.S.} = \frac{\sin \theta \cos \phi}{\sin \theta \sin \phi} - \frac{\cos \theta \sin \phi}{\sin \theta \sin \phi} \)
Simplify each term:
For the first term, \( \sin \theta \) cancels out, leaving \( \frac{\cos \phi}{\sin \phi} = \cot \phi \).
For the second term, \( \sin \phi \) cancels out, leaving \( \frac{\cos \theta}{\sin \theta} = \cot \theta \).
Therefore, \( \text{L.H.S.} = \cot \phi - \cot \theta \). This direct simplification is efficient.
Since the Right Hand Side (R.H.S.) is \( \cot \phi - \cot \theta \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the identity is proven.
In simple words: For both parts, we use the sum or difference formula for sine in the numerator. Then, we split the fraction into two smaller fractions and simplify each part. By canceling out common terms and using the definitions of cotangent and tangent, we get the expression on the right side.

🎯 Exam Tip: When proving identities involving fractions with sum/difference terms in the numerator, it's often effective to expand the numerator first and then split the fraction. This allows for direct simplification into cotangent and tangent terms.

Question 11. Prove that: \( \frac{\sin (A-B)}{\sin A \sin B}+\frac{\sin (B-C)}{\sin B \sin C}+\frac{\sin (C-A)}{\sin C \sin A} = 0 \).
Answer:
Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \frac{\sin (A-B)}{\sin A \sin B}+\frac{\sin (B-C)}{\sin B \sin C}+\frac{\sin (C-A)}{\sin C \sin A} \)
Let's expand each term separately using the difference formula for sine: \( \sin (X-Y) = \sin X \cos Y - \cos X \sin Y \).

For the first term:
\( \frac{\sin (A-B)}{\sin A \sin B} = \frac{\sin A \cos B - \cos A \sin B}{\sin A \sin B} \)
Split the fraction into two parts:
\( = \frac{\sin A \cos B}{\sin A \sin B} - \frac{\cos A \sin B}{\sin A \sin B} \)
Simplify each part:
\( = \frac{\cos B}{\sin B} - \frac{\cos A}{\sin A} = \cot B - \cot A \). This method simplifies each fraction to a difference of cotangents.

For the second term:
\( \frac{\sin (B-C)}{\sin B \sin C} = \frac{\sin B \cos C - \cos B \sin C}{\sin B \sin C} \)
Split the fraction:
\( = \frac{\sin B \cos C}{\sin B \sin C} - \frac{\cos B \sin C}{\sin B \sin C} \)
Simplify each part:
\( = \frac{\cos C}{\sin C} - \frac{\cos B}{\sin B} = \cot C - \cot B \).

For the third term:
\( \frac{\sin (C-A)}{\sin C \sin A} = \frac{\sin C \cos A - \cos C \sin A}{\sin C \sin A} \)
Split the fraction:
\( = \frac{\sin C \cos A}{\sin C \sin A} - \frac{\cos C \sin A}{\sin C \sin A} \)
Simplify each part:
\( = \frac{\cos A}{\sin A} - \frac{\cos C}{\sin C} = \cot A - \cot C \).

Now, substitute these simplified forms back into the L.H.S. equation:
\( \text{L.H.S.} = (\cot B - \cot A) + (\cot C - \cot B) + (\cot A - \cot C) \)
Rearrange the terms:
\( \text{L.H.S.} = \cot B - \cot A + \cot C - \cot B + \cot A - \cot C \)
Notice that all positive terms have a corresponding negative term:
\( (\cot B - \cot B) + (-\cot A + \cot A) + (\cot C - \cot C) \)
\( \implies \text{L.H.S.} = 0 + 0 + 0 = 0 \). The terms cancel out perfectly.
Since the Right Hand Side (R.H.S.) is 0, we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the equation is proven.
In simple words: We take each part of the sum and use the formula for sine of a difference. Then, we split each fraction into two parts and simplify them into differences of cotangents. When we add all three simplified parts together, all the cotangent terms cancel each other out, leaving us with zero.

🎯 Exam Tip: For problems involving sums of similar fractional terms, focus on simplifying one term completely first. Often, the pattern for the other terms will be similar, and they will cancel out when summed. Be careful with signs when subtracting cotangents.

Question 12. Prove that: \( \sin 105^\circ + \cos 105^\circ = \cos 45^\circ \).
Answer:
Let's start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \sin 105^\circ + \cos 105^\circ \)
We can express \( 105^\circ \) as a sum of known angles, for example, \( 60^\circ + 45^\circ \).
So, \( \text{L.H.S.} = \sin (60^\circ + 45^\circ) + \cos (60^\circ + 45^\circ) \). This makes it easier to use sum formulas.

Expand \( \sin (60^\circ + 45^\circ) \) using \( \sin (A+B) = \sin A \cos B + \cos A \sin B \):
\( \sin (60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \)
\( = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}} \).

Expand \( \cos (60^\circ + 45^\circ) \) using \( \cos (A+B) = \cos A \cos B - \sin A \sin B \):
\( \cos (60^\circ + 45^\circ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ \)
\( = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}} \).

Now, add these two expanded terms:
\( \text{L.H.S.} = \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right) + \left(\frac{1-\sqrt{3}}{2\sqrt{2}}\right) \)
Combine the numerators since they have a common denominator:
\( \implies \text{L.H.S.} = \frac{\sqrt{3}+1+1-\sqrt{3}}{2\sqrt{2}} \)
The \( \sqrt{3} \) and \( -\sqrt{3} \) terms cancel out:
\( \implies \text{L.H.S.} = \frac{1+1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} \)
Simplify the fraction:
\( \implies \text{L.H.S.} = \frac{1}{\sqrt{2}} \). This is a well-known trigonometric value.

We know that \( \cos 45^\circ = \frac{1}{\sqrt{2}} \).
So, \( \text{L.H.S.} = \cos 45^\circ \).
Since the Right Hand Side (R.H.S.) is \( \cos 45^\circ \), we have:
\( \text{L.H.S.} = \text{R.H.S.} \)
Thus, the equation is proven.
In simple words: We break down 105 degrees into 60 and 45 degrees. Then, we use the sum formulas to expand both sine and cosine terms. After adding these expanded terms, many parts cancel each other, and the expression simplifies to one over root two, which is the value of cosine 45 degrees.

🎯 Exam Tip: When given an angle like 105°, try to express it as a sum or difference of standard angles (30°, 45°, 60°, 90°) to apply the sum/difference formulas. Always remember to simplify the expression completely and rationalize denominators if needed.

Question 13. Find the value of \( \sin (\alpha + \beta), \cos (\alpha + \beta), \) and \( \tan (\alpha + \beta) \), given:
(i) \( \sin \alpha = \frac { 3 }{ 5 }, \cos \beta = \frac { 5 }{ 13 } \), \( \alpha \) and \( \beta \) in Quadrant I.
(ii) \( \cos \alpha = \frac { -12 }{ 13 }, \cot \beta = \frac { 24 }{ 7 } \), \( \alpha \) in Quadrant II, \( \beta \) in Quadrant III.
Answer:
(i) Given: \( \sin \alpha = \frac{3}{5} \), \( \cos \beta = \frac{5}{13} \), and both \( \alpha \) and \( \beta \) are in Quadrant I.
For an angle in Quadrant I, all trigonometric ratios are positive.

First, find \( \cos \alpha \):
We use the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \).
\( \cos \alpha = \sqrt{1 - \sin^2 \alpha} \) (since \( \alpha \) is in Q1, \( \cos \alpha \) is positive)
\( \implies \cos \alpha = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25-9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \).

Next, find \( \sin \beta \):
We use the identity \( \sin^2 \beta + \cos^2 \beta = 1 \).
\( \sin \beta = \sqrt{1 - \cos^2 \beta} \) (since \( \beta \) is in Q1, \( \sin \beta \) is positive)
\( \implies \sin \beta = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{169-25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \).

Now, calculate \( \sin (\alpha + \beta) \):
Using the formula \( \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
\( \implies \sin (\alpha + \beta) = \left(\frac{3}{5}\right) \left(\frac{5}{13}\right) + \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) \)
\( \implies \sin (\alpha + \beta) = \frac{15}{65} + \frac{48}{65} = \frac{15+48}{65} = \frac{63}{65} \).

Next, calculate \( \cos (\alpha + \beta) \):
Using the formula \( \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)
\( \implies \cos (\alpha + \beta) = \left(\frac{4}{5}\right) \left(\frac{5}{13}\right) - \left(\frac{3}{5}\right) \left(\frac{12}{13}\right) \)
\( \implies \cos (\alpha + \beta) = \frac{20}{65} - \frac{36}{65} = \frac{20-36}{65} = -\frac{16}{65} \).

Finally, calculate \( \tan (\alpha + \beta) \):
Using the formula \( \tan (\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} \)
\( \implies \tan (\alpha + \beta) = \frac{\frac{63}{65}}{-\frac{16}{65}} = -\frac{63}{16} \). This result aligns with \( (\alpha+\beta) \) being in Q2.

(ii) Given: \( \cos \alpha = \frac{-12}{13} \), \( \cot \beta = \frac{24}{7} \), \( \alpha \) in Quadrant II, \( \beta \) in Quadrant III.
For \( \alpha \) in Quadrant II, \( \sin \alpha \) is positive and \( \cos \alpha \) is negative. For \( \beta \) in Quadrant III, \( \sin \beta \) and \( \cos \beta \) are negative.

First, find \( \sin \alpha \):
\( \sin \alpha = \sqrt{1 - \cos^2 \alpha} \) (since \( \alpha \) is in Q2, \( \sin \alpha \) is positive)
\( \implies \sin \alpha = \sqrt{1 - \left(\frac{-12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{169-144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \).

Next, find \( \sin \beta \) and \( \cos \beta \) from \( \cot \beta = \frac{24}{7} \):
We know \( \csc^2 \beta = 1 + \cot^2 \beta \).
\( \csc \beta = -\sqrt{1 + \left(\frac{24}{7}\right)^2} \) (since \( \beta \) is in Q3, \( \csc \beta \) is negative)
\( \implies \csc \beta = -\sqrt{1 + \frac{576}{49}} = -\sqrt{\frac{49+576}{49}} = -\sqrt{\frac{625}{49}} = -\frac{25}{7} \).
So, \( \sin \beta = \frac{1}{\csc \beta} = \frac{1}{-\frac{25}{7}} = -\frac{7}{25} \).
Also, \( \cot \beta = \frac{\cos \beta}{\sin \beta} \implies \cos \beta = \cot \beta \cdot \sin \beta \).
\( \implies \cos \beta = \left(\frac{24}{7}\right) \left(-\frac{7}{25}\right) = -\frac{24}{25} \). This confirms \( \cos \beta \) is negative in Q3.

Now, calculate \( \sin (\alpha + \beta) \):
Using the formula \( \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
\( \implies \sin (\alpha + \beta) = \left(\frac{5}{13}\right) \left(-\frac{24}{25}\right) + \left(\frac{-12}{13}\right) \left(-\frac{7}{25}\right) \)
\( \implies \sin (\alpha + \beta) = -\frac{120}{325} + \frac{84}{325} = \frac{-120+84}{325} = -\frac{36}{325} \).

Next, calculate \( \cos (\alpha + \beta) \):
Using the formula \( \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)
\( \implies \cos (\alpha + \beta) = \left(\frac{-12}{13}\right) \left(-\frac{24}{25}\right) - \left(\frac{5}{13}\right) \left(-\frac{7}{25}\right) \)
\( \implies \cos (\alpha + \beta) = \frac{288}{325} - \left(-\frac{35}{325}\right) = \frac{288+35}{325} = \frac{323}{325} \).

Finally, calculate \( \tan (\alpha + \beta) \):
Using the formula \( \tan (\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} \)
\( \implies \tan (\alpha + \beta) = \frac{-\frac{36}{325}}{\frac{323}{325}} = -\frac{36}{323} \). This negative tangent is consistent with \( (\alpha+\beta) \) being in Q4, since \( \sin(\alpha+\beta) \) is negative and \( \cos(\alpha+\beta) \) is positive.
In simple words: For both parts, first find all missing sine and cosine values using the Pythagorean identity and considering the given quadrant to determine the correct sign. Then, use the angle sum formulas for sine, cosine, and tangent to calculate the required values. Pay close attention to signs based on the quadrant.

🎯 Exam Tip: Always determine the sign of the trigonometric functions based on the quadrant of the angle *before* calculating its value. Mistakes in signs are common. Clearly state which formula you are using for each step to avoid errors.

Question 14. Find the values of \( \sin (\alpha - \beta), \cos (\alpha - \beta) \) and \( \tan (\alpha - \beta) \), given:
(i) \( \sin \alpha = \frac { 8 }{ 17 }, \tan \beta = \frac { 5 }{ 12 } \), \( \alpha \) and \( \beta \) in Quadrant I.
(ii) \( \cos \alpha = \frac { -12 }{ 13 }, \cot \beta = \frac { 24 }{ 7 } \), \( \alpha \) in Quadrant II, \( \beta \) in Quadrant I.
Answer:
(i) Given: \( \sin \alpha = \frac{8}{17} \), \( \tan \beta = \frac{5}{12} \), and both \( \alpha \) and \( \beta \) are in Quadrant I.
For an angle in Quadrant I, all trigonometric ratios are positive.

First, find \( \cos \alpha \):
We use \( \cos \alpha = \sqrt{1 - \sin^2 \alpha} \) (since \( \alpha \) is in Q1, \( \cos \alpha \) is positive)
\( \implies \cos \alpha = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{289-64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17} \).

Next, find \( \sin \beta \) and \( \cos \beta \) from \( \tan \beta = \frac{5}{12} \):
We know \( \sec^2 \beta = 1 + \tan^2 \beta \).
\( \sec \beta = \sqrt{1 + \left(\frac{5}{12}\right)^2} \) (since \( \beta \) is in Q1, \( \sec \beta \) is positive)
\( \implies \sec \beta = \sqrt{1 + \frac{25}{144}} = \sqrt{\frac{144+25}{144}} = \sqrt{\frac{169}{144}} = \frac{13}{12} \).
So, \( \cos \beta = \frac{1}{\sec \beta} = \frac{1}{\frac{13}{12}} = \frac{12}{13} \).
Also, \( \tan \beta = \frac{\sin \beta}{\cos \beta} \implies \sin \beta = \tan \beta \cdot \cos \beta \).
\( \implies \sin \beta = \left(\frac{5}{12}\right) \left(\frac{12}{13}\right) = \frac{5}{13} \). This confirms \( \sin \beta \) is positive in Q1.

Now, calculate \( \sin (\alpha - \beta) \):
Using the formula \( \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \)
\( \implies \sin (\alpha - \beta) = \left(\frac{8}{17}\right) \left(\frac{12}{13}\right) - \left(\frac{15}{17}\right) \left(\frac{5}{13}\right) \)
\( \implies \sin (\alpha - \beta) = \frac{96}{221} - \frac{75}{221} = \frac{96-75}{221} = \frac{21}{221} \).

Next, calculate \( \cos (\alpha - \beta) \):
Using the formula \( \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \)
\( \implies \cos (\alpha - \beta) = \left(\frac{15}{17}\right) \left(\frac{12}{13}\right) + \left(\frac{8}{17}\right) \left(\frac{5}{13}\right) \)
\( \implies \cos (\alpha - \beta) = \frac{180}{221} + \frac{40}{221} = \frac{180+40}{221} = \frac{220}{221} \).

Finally, calculate \( \tan (\alpha - \beta) \):
Using the formula \( \tan (\alpha - \beta) = \frac{\sin (\alpha - \beta)}{\cos (\alpha - \beta)} \)
\( \implies \tan (\alpha - \beta) = \frac{\frac{21}{221}}{\frac{220}{221}} = \frac{21}{220} \).

(ii) Given: \( \cos \alpha = \frac{-12}{13} \), \( \cot \beta = \frac{24}{7} \), \( \alpha \) in Quadrant II, \( \beta \) in Quadrant I.
For \( \alpha \) in Quadrant II, \( \sin \alpha \) is positive and \( \cos \alpha \) is negative. For \( \beta \) in Quadrant I, all ratios are positive.

First, find \( \sin \alpha \):
\( \sin \alpha = \sqrt{1 - \cos^2 \alpha} \) (since \( \alpha \) is in Q2, \( \sin \alpha \) is positive)
\( \implies \sin \alpha = \sqrt{1 - \left(\frac{-12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \).

Next, find \( \sin \beta \) and \( \cos \beta \) from \( \cot \beta = \frac{24}{7} \):
We know \( \csc^2 \beta = 1 + \cot^2 \beta \).
\( \csc \beta = \sqrt{1 + \left(\frac{24}{7}\right)^2} \) (since \( \beta \) is in Q1, \( \csc \beta \) is positive)
\( \implies \csc \beta = \sqrt{1 + \frac{576}{49}} = \sqrt{\frac{49+576}{49}} = \sqrt{\frac{625}{49}} = \frac{25}{7} \).
So, \( \sin \beta = \frac{1}{\csc \beta} = \frac{1}{\frac{25}{7}} = \frac{7}{25} \).
Also, \( \cot \beta = \frac{\cos \beta}{\sin \beta} \implies \cos \beta = \cot \beta \cdot \sin \beta \).
\( \implies \cos \beta = \left(\frac{24}{7}\right) \left(\frac{7}{25}\right) = \frac{24}{25} \). This confirms \( \cos \beta \) is positive in Q1.

Now, calculate \( \sin (\alpha - \beta) \):
Using the formula \( \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \)
\( \implies \sin (\alpha - \beta) = \left(\frac{5}{13}\right) \left(\frac{24}{25}\right) - \left(\frac{-12}{13}\right) \left(\frac{7}{25}\right) \)
\( \implies \sin (\alpha - \beta) = \frac{120}{325} - \left(-\frac{84}{325}\right) = \frac{120+84}{325} = \frac{204}{325} \).

Next, calculate \( \cos (\alpha - \beta) \):
Using the formula \( \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \)
\( \implies \cos (\alpha - \beta) = \left(\frac{-12}{13}\right) \left(\frac{24}{25}\right) + \left(\frac{5}{13}\right) \left(\frac{7}{25}\right) \)
\( \implies \cos (\alpha - \beta) = -\frac{288}{325} + \frac{35}{325} = \frac{-288+35}{325} = -\frac{253}{325} \).

Finally, calculate \( \tan (\alpha - \beta) \):
Using the formula \( \tan (\alpha - \beta) = \frac{\sin (\alpha - \beta)}{\cos (\alpha - \beta)} \)
\( \implies \tan (\alpha - \beta) = \frac{\frac{204}{325}}{-\frac{253}{325}} = -\frac{204}{253} \). This negative tangent is consistent with \( (\alpha-\beta) \) being in Q2.
In simple words: For both parts, first find the missing sine or cosine values using the Pythagorean identity, making sure to apply the correct sign based on the given quadrant. Then, use the angle difference formulas for sine, cosine, and tangent to calculate the required values. Always be careful with the signs of the values you use.

🎯 Exam Tip: Accurately identifying the quadrant of each angle is critical for determining the signs of sine, cosine, and other trigonometric functions. A common mistake is using a positive square root when a negative value is required. Double-check all signs before proceeding with sum/difference calculations.

Question 21. Prove that \( 1 + \tan 2\theta \tan \theta = \sec 2\theta \).
Answer:
We start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. \( = 1 + \tan 2\theta \tan \theta \)
We can rewrite tangent functions in terms of sine and cosine:
\( = 1 + \frac{\sin 2\theta}{\cos 2\theta} \cdot \frac{\sin \theta}{\cos \theta} \)
Next, we find a common denominator for the terms:
\( = \frac{\cos 2\theta \cos \theta + \sin 2\theta \sin \theta}{\cos 2\theta \cos \theta} \)
Using the trigonometric identity \( \cos(A - B) = \cos A \cos B + \sin A \sin B \), the numerator simplifies:
\( = \frac{\cos(2\theta - \theta)}{\cos 2\theta \cos \theta} \)
\( = \frac{\cos \theta}{\cos 2\theta \cos \theta} \)
We can cancel \( \cos \theta \) from the numerator and denominator:
\( = \frac{1}{\cos 2\theta} \)
By definition, \( \frac{1}{\cos x} = \sec x \):
\( = \sec 2\theta \)
This is equal to the Right Hand Side (R.H.S.) of the equation. So, the identity is proven.
In simple words: To prove this, we change the tangent parts into sine and cosine. Then, we use a special math rule for cosines to simplify the top part of the fraction. After canceling some terms, we are left with the secant of two theta.

🎯 Exam Tip: When proving trigonometric identities, convert all tangent and cotangent terms to sine and cosine first. Look for opportunities to apply sum/difference or double angle formulas to simplify expressions.

Question 22. Prove that \( 4 \sin \left(\frac{\pi}{3}-\theta\right) \sin \left(\frac{\pi}{3}+\theta\right) = 3 – 4 \sin^2\theta \).
Answer:
Let's start with the Left Hand Side (L.H.S.):
L.H.S. \( = 4 \sin \left(\frac{\pi}{3}-\theta\right) \sin \left(\frac{\pi}{3}+\theta\right) \)
We use the trigonometric identity \( \sin(A-B)\sin(A+B) = \sin^2 A - \sin^2 B \). Here, \( A = \frac{\pi}{3} \) and \( B = \theta \).
\( = 4 \left[ \sin^2 \left(\frac{\pi}{3}\right) - \sin^2 \theta \right] \)
We know that \( \sin \left(\frac{\pi}{3}\right) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \).
\( = 4 \left[ \left(\frac{\sqrt{3}}{2}\right)^2 - \sin^2 \theta \right] \)
\( = 4 \left[ \frac{3}{4} - \sin^2 \theta \right] \)
Now, we multiply 4 into the bracket:
\( = 4 \times \frac{3}{4} - 4 \sin^2 \theta \)
\( = 3 - 4 \sin^2 \theta \)
This matches the Right Hand Side (R.H.S.) of the equation. The identity is proven.
In simple words: We used a special formula for multiplying sines with sums and differences. By replacing \( \sin(60^\circ) \) with its value and simplifying, we get the answer \( 3 - 4 \sin^2\theta \).

🎯 Exam Tip: Memorizing product-to-sum and sum-to-product trigonometric identities (like \( \sin(A-B)\sin(A+B) = \sin^2 A - \sin^2 B \)) can save a lot of time in proofs.

Question 23. Prove that \( \frac{\cos 17^\circ + \sin 17^\circ}{\cos 17^\circ - \sin 17^\circ} = \tan 62^\circ \).
Answer:
Let's start with the Left Hand Side (L.H.S.):
L.H.S. \( = \frac{\cos 17^\circ + \sin 17^\circ}{\cos 17^\circ - \sin 17^\circ} \)
To simplify this expression, we divide both the numerator and the denominator by \( \cos 17^\circ \):
\( = \frac{\frac{\cos 17^\circ}{\cos 17^\circ} + \frac{\sin 17^\circ}{\cos 17^\circ}}{\frac{\cos 17^\circ}{\cos 17^\circ} - \frac{\sin 17^\circ}{\cos 17^\circ}} \)
This simplifies to:
\( = \frac{1 + \tan 17^\circ}{1 - \tan 17^\circ} \)
We know that \( \tan 45^\circ = 1 \). So, we can replace 1 in the numerator and the denominator with \( \tan 45^\circ \):
\( = \frac{\tan 45^\circ + \tan 17^\circ}{1 - \tan 45^\circ \tan 17^\circ} \)
This expression matches the tangent addition formula \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). Here, \( A = 45^\circ \) and \( B = 17^\circ \).
\( = \tan(45^\circ + 17^\circ) \)
\( = \tan 62^\circ \)
This is equal to the Right Hand Side (R.H.S.) of the equation. The identity is proven.
In simple words: We changed the expression by dividing everything by \( \cos 17^\circ \). This turned it into a form of \( \frac{1+\tan B}{1-\tan B} \). Knowing that \( 1 = \tan 45^\circ \), we used a tangent addition rule to combine \( 45^\circ \) and \( 17^\circ \) which results in \( \tan 62^\circ \).

🎯 Exam Tip: When you see expressions like \( \frac{\cos A + \sin A}{\cos A - \sin A} \), always think of dividing by \( \cos A \) to get \( \frac{1 + \tan A}{1 - \tan A} \), which can then be written as \( \tan(45^\circ + A) \).

Question 24. Prove that \( \tan 3A - \tan 2A - \tan A = \tan 3A \tan 2A \tan A \).
Answer:
We know that \( 3A = 2A + A \). Let's take the tangent of both sides:
\( \tan(3A) = \tan(2A + A) \)
Using the tangent addition formula \( \tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y} \), with \( X = 2A \) and \( Y = A \):
\( \tan 3A = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A} \)
Now, multiply both sides by \( (1 - \tan 2A \tan A) \):
\( \tan 3A (1 - \tan 2A \tan A) = \tan 2A + \tan A \)
Distribute \( \tan 3A \) on the left side:
\( \tan 3A - \tan 3A \tan 2A \tan A = \tan 2A + \tan A \)
Finally, rearrange the terms to match the required proof. Move \( \tan 3A \tan 2A \tan A \) to the right side and \( \tan 2A + \tan A \) to the left side:
\( \tan 3A - \tan 2A - \tan A = \tan 3A \tan 2A \tan A \)
The identity is proven.
In simple words: We started by writing \( \tan 3A \) as \( \tan(2A+A) \). Then, we used the rule for adding tangents. After that, we just rearranged the equation to get the final answer. This identity helps simplify expressions involving multiple angles.

🎯 Exam Tip: For identities involving three angles (like A, 2A, 3A), a common strategy is to express the largest angle as the sum of the other two (e.g., \( 3A = 2A + A \)) and then apply the tangent addition formula.

Question 25. Prove that \( \tan 75^\circ - \tan 30^\circ - \tan 75^\circ \tan 30^\circ = 1 \).
Answer:
This problem asks us to prove an identity that looks like a rearranged form of the tangent addition or subtraction formula. Let's start with the tangent of a sum or difference that relates to \( 75^\circ \) and \( 30^\circ \). We know \( 75^\circ = 45^\circ + 30^\circ \).
Consider the formula for \( \tan(A+B) \):
\( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Let \( A = 45^\circ \) and \( B = 30^\circ \). Then \( A+B = 75^\circ \).
So, \( \tan 75^\circ = \tan(45^\circ + 30^\circ) \)
\( \tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \)
We know that \( \tan 45^\circ = 1 \). Substitute this value into the equation:
\( \tan 75^\circ = \frac{1 + \tan 30^\circ}{1 - 1 \cdot \tan 30^\circ} \)
\( \tan 75^\circ = \frac{1 + \tan 30^\circ}{1 - \tan 30^\circ} \)
Now, multiply both sides by \( (1 - \tan 30^\circ) \):
\( \tan 75^\circ (1 - \tan 30^\circ) = 1 + \tan 30^\circ \)
Distribute \( \tan 75^\circ \) on the left side:
\( \tan 75^\circ - \tan 75^\circ \tan 30^\circ = 1 + \tan 30^\circ \)
Finally, rearrange the terms to match the required proof. Move \( \tan 30^\circ \) from the right side to the left side:
\( \tan 75^\circ - \tan 30^\circ - \tan 75^\circ \tan 30^\circ = 1 \)
The identity is proven.
In simple words: We started with the tangent addition formula for \( \tan(45^\circ + 30^\circ) \). After putting in the value of \( \tan 45^\circ \), we rearranged the equation. This led us straight to the result we needed to prove.

🎯 Exam Tip: Recognize that the given equation is a rearrangement of the \( \tan(A+B) \) formula where \( A+B = 45^\circ + 30^\circ = 75^\circ \). Start with the standard formula and work backward to prove the specific rearrangement.

Question 26. Prove that \( \cos 2\theta \cos 2\Phi + \sin^2 (\theta – \Phi) – \sin^2 (\theta + \Phi) = \cos (2\theta + 2\Phi) \).
Answer:
Let's start with the Left Hand Side (L.H.S.):
L.H.S. \( = \cos 2\theta \cos 2\Phi + \sin^2 (\theta – \Phi) – \sin^2 (\theta + \Phi) \)
We use the trigonometric identity \( \sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B) \).
Here, let \( A = (\theta - \Phi) \) and \( B = (\theta + \Phi) \).
Then, \( \sin^2 (\theta – \Phi) – \sin^2 (\theta + \Phi) = \sin \left[ (\theta - \Phi) + (\theta + \Phi) \right] \sin \left[ (\theta - \Phi) - (\theta + \Phi) \right] \)
Simplify the arguments of the sine functions:
\( = \sin (\theta - \Phi + \theta + \Phi) \sin (\theta - \Phi - \theta - \Phi) \)
\( = \sin (2\theta) \sin (-2\Phi) \)
Since \( \sin(-x) = -\sin x \):
\( = -\sin (2\theta) \sin (2\Phi) \)
Now, substitute this back into the L.H.S. expression:
L.H.S. \( = \cos 2\theta \cos 2\Phi - \sin 2\theta \sin 2\Phi \)
This expression matches the cosine addition formula \( \cos(X+Y) = \cos X \cos Y - \sin X \sin Y \).
Here, let \( X = 2\theta \) and \( Y = 2\Phi \).
\( = \cos (2\theta + 2\Phi) \)
This is equal to the Right Hand Side (R.H.S.) of the equation. The identity is proven.
In simple words: We used a special formula to simplify the \( \sin^2 \) parts. This changed the expression into a form like \( \cos A \cos B - \sin A \sin B \), which is the rule for \( \cos(A+B) \). By doing this, we got the answer \( \cos(2\theta + 2\Phi) \).

🎯 Exam Tip: Always look for common trigonometric identities like the difference of squares of sines (which becomes a product of sines) to simplify complex terms before trying to combine everything.

Question 27. If \( \sin (\theta + \alpha) = \cos (\theta + \alpha) \), prove that \( \tan \theta = \frac{1-\tan \alpha}{1+\tan \alpha} \).
Answer:
We are given the initial condition:
\( \sin (\theta + \alpha) = \cos (\theta + \alpha) \)
To work with tangent functions, we can divide both sides by \( \cos (\theta + \alpha) \). This is valid as long as \( \cos (\theta + \alpha) \neq 0 \).
\( \frac{\sin (\theta + \alpha)}{\cos (\theta + \alpha)} = 1 \)
This simplifies to:
\( \tan (\theta + \alpha) = 1 \)
We know that \( \tan 45^\circ = 1 \) or \( \tan \frac{\pi}{4} = 1 \). So, we can write:
\( \theta + \alpha = \frac{\pi}{4} \)
Now, we want to find \( \tan \theta \), so let's isolate \( \theta \):
\( \theta = \frac{\pi}{4} - \alpha \)
Take the tangent of both sides of this equation:
\( \tan \theta = \tan \left(\frac{\pi}{4} - \alpha\right) \)
Using the tangent subtraction formula \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \), with \( A = \frac{\pi}{4} \) and \( B = \alpha \):
\( \tan \theta = \frac{\tan \frac{\pi}{4} - \tan \alpha}{1 + \tan \frac{\pi}{4} \tan \alpha} \)
Substitute \( \tan \frac{\pi}{4} = 1 \):
\( \tan \theta = \frac{1 - \tan \alpha}{1 + 1 \cdot \tan \alpha} \)
\( \tan \theta = \frac{1 - \tan \alpha}{1 + \tan \alpha} \)
The expression is proven.
In simple words: First, we changed the given sine and cosine equation into a tangent equation. This showed us that \( \theta + \alpha \) must be \( 45^\circ \). Then, we moved \( \alpha \) to the other side and applied the tangent function to both sides. Using the tangent subtraction rule for \( 45^\circ \), we found the desired result.

🎯 Exam Tip: When \( \sin X = \cos X \), remember that \( \tan X = 1 \). This simplifies the problem significantly, allowing you to use the standard tangent sum/difference formulas efficiently.

Question 28. Given that \( A = B + C \), prove that \( \tan A - \tan B - \tan C = \tan A \tan B \tan C \).
Answer:
We are given the relationship between the angles:
\( A = B + C \)
To involve tangent functions, let's take the tangent of both sides of this equation:
\( \tan A = \tan (B + C) \)
Now, apply the tangent addition formula \( \tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y} \), with \( X = B \) and \( Y = C \):
\( \tan A = \frac{\tan B + \tan C}{1 - \tan B \tan C} \)
To remove the fraction, multiply both sides by \( (1 - \tan B \tan C) \):
\( \tan A (1 - \tan B \tan C) = \tan B + \tan C \)
Distribute \( \tan A \) on the left side:
\( \tan A - \tan A \tan B \tan C = \tan B + \tan C \)
Finally, rearrange the terms to match the required proof. We want all tangent terms on the left side except for the product \( \tan A \tan B \tan C \), which should be on the right side. Move \( \tan B \) and \( \tan C \) to the left, and \( \tan A \tan B \tan C \) to the right:
\( \tan A - \tan B - \tan C = \tan A \tan B \tan C \)
The identity is proven.
In simple words: We started with the given angle relation \( A = B + C \) and took the tangent of both sides. By using the tangent addition formula and then rearranging the terms, we arrived at the required identity. This is a common way to relate tangents of angles in a sum.

🎯 Exam Tip: This identity is a common result when \( A = B+C \). Remember to start by taking the tangent of both sides of the angle relationship and expand using the sum formula.

Question 29. If \( A + B = 45^\circ \), show that \( \tan A + \tan B + \tan A \tan B = 1 \). Hence, or otherwise, express \( \tan 22^\circ30' \) in surd form.
Answer:
**Part 1: Show that \( \tan A + \tan B + \tan A \tan B = 1 \)**
We are given that \( A + B = 45^\circ \).
Take the tangent of both sides:
\( \tan(A + B) = \tan 45^\circ \)
Using the tangent addition formula \( \tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y} \):
\( \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1 \)
Multiply both sides by \( (1 - \tan A \tan B) \):
\( \tan A + \tan B = 1 - \tan A \tan B \)
Now, move the term \( -\tan A \tan B \) to the left side:
\( \tan A + \tan B + \tan A \tan B = 1 \)
This identity is proven.

**Part 2: Express \( \tan 22^\circ30' \) in surd form**
Let \( \theta = 22^\circ30' \). We want to find \( \tan \theta \).
Notice that \( \theta + \theta = 22^\circ30' + 22^\circ30' = 45^\circ \).
So, we can use the identity we just proved by setting \( A = \theta \) and \( B = \theta \).
Substitute \( \tan A = \tan \theta \) and \( \tan B = \tan \theta \) into the identity \( \tan A + \tan B + \tan A \tan B = 1 \):
\( \tan \theta + \tan \theta + \tan \theta \cdot \tan \theta = 1 \)
\( 2 \tan \theta + \tan^2 \theta = 1 \)
Rearrange this into a quadratic equation in terms of \( \tan \theta \):
\( \tan^2 \theta + 2 \tan \theta - 1 = 0 \)
Let \( x = \tan \theta \). The equation becomes \( x^2 + 2x - 1 = 0 \).
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} \)
\( x = \frac{-2 \pm \sqrt{4 + 4}}{2} \)
\( x = \frac{-2 \pm \sqrt{8}}{2} \)
\( x = \frac{-2 \pm 2\sqrt{2}}{2} \)
\( x = -1 \pm \sqrt{2} \)
Since \( 22^\circ30' \) is in the first quadrant (between \( 0^\circ \) and \( 90^\circ \)), its tangent value must be positive. Therefore, we choose the positive root.
\( \tan 22^\circ30' = -1 + \sqrt{2} \)
Or, usually written as:
\( \tan 22^\circ30' = \sqrt{2} - 1 \)
In simple words: First, we used the angle sum \( A+B=45^\circ \) with the tangent formula to prove a new identity. Then, we used this identity by letting both \( A \) and \( B \) be \( 22^\circ30' \). This gave us a quadratic equation for \( \tan 22^\circ30' \). Solving it, and picking the positive answer since \( 22^\circ30' \) is a first quadrant angle, we got \( \sqrt{2} - 1 \).

🎯 Exam Tip: The identity \( \tan A + \tan B + \tan A \tan B = 1 \) is very useful for problems where \( A+B = 45^\circ \). For quadratic equations, always check the quadrant of the angle to pick the correct sign for the trigonometric value.

Question 30.
(i) Use the expansion of \( \tan (A – B) \) to find \( \tan 15^\circ \) without the use of tables, leaving your answer in surd form with an integral denominator.
(ii) Prove that \( \frac{2 \tan A}{1+\tan^2 A} = \sin 2A \).
Answer:
**(i) Find \( \tan 15^\circ \)**
We can write \( 15^\circ \) as the difference of two standard angles: \( 15^\circ = 45^\circ - 30^\circ \).
Using the tangent subtraction formula \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \), with \( A = 45^\circ \) and \( B = 30^\circ \):
\( \tan 15^\circ = \tan(45^\circ - 30^\circ) \)
\( = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \)
Substitute the known values \( \tan 45^\circ = 1 \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\( = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \)
To simplify, find a common denominator in the numerator and denominator:
\( = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} \)
Cancel \( \sqrt{3} \) from the numerator and denominator:
\( = \frac{\sqrt{3}-1}{\sqrt{3}+1} \)
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is \( (\sqrt{3}-1) \):
\( = \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \)
\( = \frac{(\sqrt{3})^2 - 2\sqrt{3}(1) + 1^2}{(\sqrt{3})^2 - 1^2} \)
\( = \frac{3 - 2\sqrt{3} + 1}{3 - 1} \)
\( = \frac{4 - 2\sqrt{3}}{2} \)
Divide both terms in the numerator by 2:
\( = 2 - \sqrt{3} \)

**(ii) Prove that \( \frac{2 \tan A}{1+\tan^2 A} = \sin 2A \)**
Let's start with the Left Hand Side (L.H.S.):
L.H.S. \( = \frac{2 \tan A}{1+\tan^2 A} \)
We know the trigonometric identity \( 1+\tan^2 A = \sec^2 A \). Substitute this into the denominator:
\( = \frac{2 \tan A}{\sec^2 A} \)
Now, express \( \tan A \) as \( \frac{\sin A}{\cos A} \) and \( \sec^2 A \) as \( \frac{1}{\cos^2 A} \):
\( = \frac{2 \left(\frac{\sin A}{\cos A}\right)}{\frac{1}{\cos^2 A}} \)
To simplify, multiply by the reciprocal of the denominator:
\( = 2 \left(\frac{\sin A}{\cos A}\right) \cdot (\cos^2 A) \)
\( = 2 \sin A \cos A \)
This is the double-angle formula for sine, which is equal to \( \sin 2A \).
\( = \sin 2A \)
This matches the Right Hand Side (R.H.S.) of the equation. The identity is proven.
In simple words: For \( \tan 15^\circ \), we split \( 15^\circ \) into \( 45^\circ - 30^\circ \) and used the tangent subtraction rule. After putting in values and simplifying, we got \( 2 - \sqrt{3} \). For the proof, we changed \( \tan A \) and \( \sec A \) into sine and cosine. This simplified the expression to \( 2 \sin A \cos A \), which is the formula for \( \sin 2A \).

🎯 Exam Tip: For problems like \( \tan 15^\circ \), \( \tan 75^\circ \), etc., always express the angle as a sum or difference of standard angles like \( 30^\circ, 45^\circ, 60^\circ \). For trigonometric identities, converting to sine and cosine is a reliable strategy if no other identity immediately applies.

Question 31. If \( A + B = 225^\circ \), prove that \( \tan A + \tan B = 1 – \tan A \tan B \).
Answer:
We are given the sum of two angles:
\( A + B = 225^\circ \)
To introduce tangent functions, take the tangent of both sides of the equation:
\( \tan(A + B) = \tan 225^\circ \)
We need to find the value of \( \tan 225^\circ \). We can write \( 225^\circ \) as \( 180^\circ + 45^\circ \).
Using the identity \( \tan(180^\circ + x) = \tan x \):
\( \tan 225^\circ = \tan (180^\circ + 45^\circ) = \tan 45^\circ \)
We know that \( \tan 45^\circ = 1 \).
So, the equation becomes:
\( \tan(A + B) = 1 \)
Now, apply the tangent addition formula \( \tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y} \):
\( \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1 \)
To eliminate the fraction, multiply both sides by \( (1 - \tan A \tan B) \):
\( \tan A + \tan B = 1 \cdot (1 - \tan A \tan B) \)
\( \tan A + \tan B = 1 - \tan A \tan B \)
The identity is proven.
In simple words: We started with the given sum of angles, \( A+B=225^\circ \). We found that \( \tan 225^\circ \) is equal to 1. Then, we used the rule for adding tangents, \( \tan(A+B) \), and set it equal to 1. Rearranging this equation gave us the answer we needed to prove.

🎯 Exam Tip: When dealing with angles greater than \( 90^\circ \), always reduce them to acute angles using the properties of trigonometric functions in different quadrants (e.g., \( \tan(180^\circ + x) = \tan x \)) before applying sum/difference formulas.