Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.5

Question 1. Find the value of cos 2A, A lies in the first quadrant when
(i) \( \cos A = \frac{15}{17} \)
(ii) \( \sin A = \frac{4}{5} \)
(iii) \( \tan A = \frac{16}{63} \)
Answer:
(i) Given \( \cos A = \frac{15}{17} \)
We know the identity: \( \sin^2 A + \cos^2 A = 1 \)
So, \( \sin^2 A = 1 - \cos^2 A \)
\( \sin^2 A = 1 - \left(\frac{15}{17}\right)^2 \)
\( \sin^2 A = 1 - \frac{225}{289} \)
\( \sin^2 A = \frac{289 - 225}{289} \)
\( \sin^2 A = \frac{64}{289} \)
\( \sin A = \pm \sqrt{\frac{64}{289}} \)
\( \sin A = \pm \frac{8}{17} \)
Since A is in the first quadrant, \( \sin A \) must be positive.
So, \( \sin A = \frac{8}{17} \)
Now, we find \( \cos 2A \) using the formula \( \cos 2A = \cos^2 A - \sin^2 A \)
\( \cos 2A = \left(\frac{15}{17}\right)^2 - \left(\frac{8}{17}\right)^2 \)
\( \cos 2A = \frac{225}{289} - \frac{64}{289} \)
\( \cos 2A = \frac{225 - 64}{289} \)
\( \cos 2A = \frac{161}{289} \)
(ii) Given \( \sin A = \frac{4}{5} \)
We know the identity: \( \sin^2 A + \cos^2 A = 1 \)
So, \( \cos^2 A = 1 - \sin^2 A \)
\( \cos^2 A = 1 - \left(\frac{4}{5}\right)^2 \)
\( \cos^2 A = 1 - \frac{16}{25} \)
\( \cos^2 A = \frac{25 - 16}{25} \)
\( \cos^2 A = \frac{9}{25} \)
\( \cos A = \pm \sqrt{\frac{9}{25}} \)
\( \cos A = \pm \frac{3}{5} \)
Since A is in the first quadrant, \( \cos A \) must be positive.
So, \( \cos A = \frac{3}{5} \)
Now, we find \( \cos 2A \) using the formula \( \cos 2A = \cos^2 A - \sin^2 A \)
\( \cos 2A = \left(\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 \)
\( \cos 2A = \frac{9}{25} - \frac{16}{25} \)
\( \cos 2A = \frac{9 - 16}{25} \)
\( \cos 2A = \frac{-7}{25} \)
(iii) Given \( \tan A = \frac{16}{63} \)
We use the formula for \( \cos 2A \) in terms of \( \tan A \):
\( \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} \)
\( \cos 2A = \frac{1 - \left(\frac{16}{63}\right)^2}{1 + \left(\frac{16}{63}\right)^2} \)
\( \cos 2A = \frac{1 - \frac{256}{3969}}{1 + \frac{256}{3969}} \)
\( \cos 2A = \frac{\frac{3969 - 256}{3969}}{\frac{3969 + 256}{3969}} \)
\( \cos 2A = \frac{3969 - 256}{3969 + 256} \)
\( \cos 2A = \frac{3713}{4225} \)
In simple words: To find \( \cos 2A \), we use a suitable trigonometric formula. If we know \( \cos A \) or \( \sin A \), we use \( \cos 2A = \cos^2 A - \sin^2 A \). If we know \( \tan A \), we use \( \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} \). Remember that the quadrant of A helps decide if \( \sin A \) or \( \cos A \) are positive or negative.

🎯 Exam Tip: Always remember the different formulas for \( \cos 2A \) and choose the one that directly uses the given information, like \( \cos^2 A - \sin^2 A \) or \( \frac{1 - \tan^2 A}{1 + \tan^2 A} \).

Question 2. If \( \theta \) be an acute angle, find
(i) \( \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \), when \( \sin \theta = \frac{1}{25} \)
(ii) \( \cos \left(\frac{\pi}{4}+\frac{\theta}{2}\right) \), when \( \sin \theta = \frac{8}{9} \)
Answer:
(i) We need to find \( \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \). We use the half-angle formula for sine, but first rewrite the angle.
We know that \( \sin \left(\frac{X}{2}\right) = \sqrt{\frac{1 - \cos X}{2}} \). Here, let \( X = \frac{\pi}{2} - \theta \).
So, \( \frac{X}{2} = \frac{1}{2}\left(\frac{\pi}{2} - \theta\right) = \frac{\pi}{4} - \frac{\theta}{2} \).
Then \( \cos X = \cos \left(\frac{\pi}{2} - \theta\right) \). We know \( \cos \left(\frac{\pi}{2} - \theta\right) = \sin \theta \).
Thus, \( \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) = \sqrt{\frac{1 - \sin \theta}{2}} \)
Given \( \sin \theta = \frac{1}{25} \). Substitute this value:
\( \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) = \sqrt{\frac{1 - \frac{1}{25}}{2}} \)
\( = \sqrt{\frac{\frac{25-1}{25}}{2}} \)
\( = \sqrt{\frac{\frac{24}{25}}{2}} \)
\( = \sqrt{\frac{24}{50}} \)
\( = \sqrt{\frac{12}{25}} \)
\( = \frac{\sqrt{12}}{\sqrt{25}} \)
\( = \frac{2\sqrt{3}}{5} \)
(ii) We need to find \( \cos \left(\frac{\pi}{4}+\frac{\theta}{2}\right) \). We use the half-angle formula for cosine.
We know that \( \cos \left(\frac{X}{2}\right) = \sqrt{\frac{1 + \cos X}{2}} \). Here, let \( X = \frac{\pi}{2} + \theta \).
So, \( \frac{X}{2} = \frac{1}{2}\left(\frac{\pi}{2} + \theta\right) = \frac{\pi}{4} + \frac{\theta}{2} \).
Then \( \cos X = \cos \left(\frac{\pi}{2} + \theta\right) \). We know \( \cos \left(\frac{\pi}{2} + \theta\right) = -\sin \theta \).
Thus, \( \cos \left(\frac{\pi}{4}+\frac{\theta}{2}\right) = \sqrt{\frac{1 - \sin \theta}{2}} \)
Given \( \sin \theta = \frac{8}{9} \). Substitute this value:
\( \cos \left(\frac{\pi}{4}+\frac{\theta}{2}\right) = \sqrt{\frac{1 - \frac{8}{9}}{2}} \)
\( = \sqrt{\frac{\frac{9-8}{9}}{2}} \)
\( = \sqrt{\frac{\frac{1}{9}}{2}} \)
\( = \sqrt{\frac{1}{18}} \)
\( = \frac{1}{\sqrt{18}} \)
\( = \frac{1}{3\sqrt{2}} \)
In simple words: To solve these problems, we use half-angle formulas. We convert the angle into a form like \( X/2 \) and then use the correct half-angle formula that relates it to \( \sin X \) or \( \cos X \). Then, we simplify the square root.

🎯 Exam Tip: Remember the basic trigonometric identities and half-angle formulas like \( \sin^2 \frac{X}{2} = \frac{1 - \cos X}{2} \) and \( \cos^2 \frac{X}{2} = \frac{1 + \cos X}{2} \). These help simplify complex angles.

Question 3. If \( \cos \theta =\frac{1}{2}\left(a+\frac{1}{a}\right) \), show that \( \cos 3\theta = \frac{1}{2}\left(a^{3}+\frac{1}{a^{3}}\right) \)
Answer:
We are given \( \cos \theta = \frac{1}{2}\left(a+\frac{1}{a}\right) \).
We need to prove \( \cos 3\theta = \frac{1}{2}\left(a^3+\frac{1}{a^3}\right) \).
We know the identity for \( \cos 3\theta \):
\( \cos 3\theta = 4\cos^3\theta - 3\cos\theta \)
Substitute the given value of \( \cos \theta \) into the identity:
\( \cos 3\theta = 4 \left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right]^3 - 3 \left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right] \)
First, calculate the cube term:
\( \left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right]^3 = \frac{1}{8}\left(a+\frac{1}{a}\right)^3 \)
Using the expansion \( (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \):
\( \left(a+\frac{1}{a}\right)^3 = a^3 + 3a^2\left(\frac{1}{a}\right) + 3a\left(\frac{1}{a}\right)^2 + \left(\frac{1}{a}\right)^3 \)
\( = a^3 + 3a + \frac{3}{a} + \frac{1}{a^3} \)
Now substitute this back into the expression for \( \cos 3\theta \):
\( \cos 3\theta = 4 \cdot \frac{1}{8}\left(a^3 + 3a + \frac{3}{a} + \frac{1}{a^3}\right) - \frac{3}{2}\left(a+\frac{1}{a}\right) \)
\( \cos 3\theta = \frac{1}{2}\left(a^3 + 3a + \frac{3}{a} + \frac{1}{a^3}\right) - \frac{3}{2}\left(a+\frac{1}{a}\right) \)
Now, combine the terms by factoring out \( \frac{1}{2} \):
\( \cos 3\theta = \frac{1}{2} \left[ \left(a^3 + 3a + \frac{3}{a} + \frac{1}{a^3}\right) - 3\left(a+\frac{1}{a}\right) \right] \)
\( \cos 3\theta = \frac{1}{2} \left[ a^3 + 3a + \frac{3}{a} + \frac{1}{a^3} - 3a - \frac{3}{a} \right] \)
Cancel the `3a` and `\frac{3}{a}` terms:
\( \cos 3\theta = \frac{1}{2} \left[ a^3 + \frac{1}{a^3} \right] \)
This proves the desired result.
In simple words: We used the formula for \( \cos 3\theta \) and put the given value of \( \cos \theta \) into it. Then, we expanded the terms carefully. After simplifying, many parts cancelled out, leaving us with the expression we wanted to prove. This shows how algebra and trigonometry work together.

🎯 Exam Tip: When proving identities, always start with the more complex side or the known identity and simplify it using given information. Be careful with algebraic expansions and cancellations.

Question 4. Prove that \( \cos 5\theta = 16 \cos^5\theta – 20 \cos^3\theta + 5 \cos \theta \)
Answer:
We need to prove the identity \( \cos 5\theta = 16 \cos^5\theta – 20 \cos^3\theta + 5 \cos \theta \).
Start by expressing \( \cos 5\theta \) as a sum of angles:
\( \cos 5\theta = \cos (3\theta + 2\theta) \)
Using the identity \( \cos(A+B) = \cos A \cos B - \sin A \sin B \):
\( \cos 5\theta = \cos 3\theta \cos 2\theta - \sin 3\theta \sin 2\theta \)
Now, substitute the formulas for \( \cos 3\theta \), \( \cos 2\theta \), \( \sin 3\theta \), and \( \sin 2\theta \) in terms of \( \cos \theta \) and \( \sin \theta \):
\( \cos 3\theta = 4\cos^3\theta - 3\cos\theta \)
\( \cos 2\theta = 2\cos^2\theta - 1 \)
\( \sin 3\theta = 3\sin\theta - 4\sin^3\theta = \sin\theta (3 - 4\sin^2\theta) \)
\( \sin 2\theta = 2\sin\theta\cos\theta \)
Substitute these into the expression for \( \cos 5\theta \):
\( \cos 5\theta = (4\cos^3\theta - 3\cos\theta)(2\cos^2\theta - 1) - [\sin\theta (3 - 4\sin^2\theta)](2\sin\theta\cos\theta) \)
Expand the first term:
\( (4\cos^3\theta - 3\cos\theta)(2\cos^2\theta - 1) = 8\cos^5\theta - 4\cos^3\theta - 6\cos^3\theta + 3\cos\theta \)
\( = 8\cos^5\theta - 10\cos^3\theta + 3\cos\theta \) (Equation 1)
Now, expand the second term. First, replace \( \sin^2\theta \) with \( 1-\cos^2\theta \):
\( [\sin\theta (3 - 4\sin^2\theta)](2\sin\theta\cos\theta) = 2\sin^2\theta\cos\theta (3 - 4\sin^2\theta) \)
\( = 2(1-\cos^2\theta)\cos\theta (3 - 4(1-\cos^2\theta)) \)
\( = 2\cos\theta(1-\cos^2\theta) (3 - 4 + 4\cos^2\theta) \)
\( = 2\cos\theta(1-\cos^2\theta) (4\cos^2\theta - 1) \)
\( = 2\cos\theta (4\cos^2\theta - 1 - 4\cos^4\theta + \cos^2\theta) \)
\( = 2\cos\theta (5\cos^2\theta - 4\cos^4\theta - 1) \)
\( = 10\cos^3\theta - 8\cos^5\theta - 2\cos\theta \) (Equation 2)
Now, subtract Equation 2 from Equation 1:
\( \cos 5\theta = (8\cos^5\theta - 10\cos^3\theta + 3\cos\theta) - (10\cos^3\theta - 8\cos^5\theta - 2\cos\theta) \)
\( \cos 5\theta = 8\cos^5\theta - 10\cos^3\theta + 3\cos\theta - 10\cos^3\theta + 8\cos^5\theta + 2\cos\theta \)
Combine like terms:
\( \cos 5\theta = (8\cos^5\theta + 8\cos^5\theta) + (-10\cos^3\theta - 10\cos^3\theta) + (3\cos\theta + 2\cos\theta) \)
\( \cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta \)
Thus, the identity is proven.
In simple words: To prove this, we broke \( \cos 5\theta \) into \( \cos(3\theta + 2\theta) \) and used the addition formula. Then we replaced \( \cos 3\theta, \cos 2\theta, \sin 3\theta, \sin 2\theta \) with their equivalent expressions in terms of \( \cos \theta \) and \( \sin \theta \). After expanding everything and simplifying, all \( \sin \theta \) terms disappeared, and we were left only with \( \cos \theta \) terms, which matched the target equation.

🎯 Exam Tip: When working with higher multiple angles like \( \cos 5\theta \), it's often helpful to break them down into sums or differences of known multiple angles (e.g., \( 3\theta + 2\theta \)) and use the addition/subtraction formulas repeatedly.

Question 5. Prove that \( \sin 4\alpha = 4 \tan \alpha \frac{1-\tan ^{2} \alpha}{\left(1+\tan ^{2} \alpha\right)^{2}} \)
Answer:
We need to prove the identity \( \sin 4\alpha = 4 \tan \alpha \frac{1-\tan ^{2} \alpha}{\left(1+\tan ^{2} \alpha\right)^{2}} \).
We will start from the left-hand side, \( \sin 4\alpha \).
We know the double angle formula \( \sin 2X = 2 \sin X \cos X \). Let \( X = 2\alpha \):
\( \sin 4\alpha = 2 \sin 2\alpha \cos 2\alpha \)
Now, we replace \( \sin 2\alpha \) and \( \cos 2\alpha \) with their formulas in terms of \( \tan \alpha \):
\( \sin 2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} \)
\( \cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} \)
Substitute these into the expression for \( \sin 4\alpha \):
\( \sin 4\alpha = 2 \left( \frac{2 \tan \alpha}{1 + \tan^2 \alpha} \right) \left( \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} \right) \)
Multiply the terms:
\( \sin 4\alpha = \frac{2 \cdot (2 \tan \alpha) \cdot (1 - \tan^2 \alpha)}{(1 + \tan^2 \alpha) \cdot (1 + \tan^2 \alpha)} \)
\( \sin 4\alpha = \frac{4 \tan \alpha (1 - \tan^2 \alpha)}{(1 + \tan^2 \alpha)^2} \)
This matches the right-hand side of the identity, so it is proven.
In simple words: We used the double angle formulas to break down \( \sin 4\alpha \) step-by-step. First, we changed \( \sin 4\alpha \) to \( 2 \sin 2\alpha \cos 2\alpha \). Then, we replaced \( \sin 2\alpha \) and \( \cos 2\alpha \) with their forms using \( \tan \alpha \). Multiplying these together gave us the final expression we needed to prove.

🎯 Exam Tip: When proving identities involving \( \sin \) and \( \cos \) with tangent in the final expression, always convert \( \sin 2A \) and \( \cos 2A \) to their \( \tan A \) forms. This often simplifies the proof significantly.

Question 6. If \( A + B = 45^\circ \), show that \( (1 + \tan A) (1 + \tan B) = 2 \)
Answer:
We are given that \( A + B = 45^\circ \).
Take the tangent of both sides of the equation:
\( \tan(A + B) = \tan 45^\circ \)
We know that \( \tan 45^\circ = 1 \).
Using the tangent addition formula \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \):
\( \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1 \)
Multiply both sides by \( (1 - \tan A \tan B) \):
\( \tan A + \tan B = 1 - \tan A \tan B \) (Equation 1)
Now, let's expand the expression we need to prove:
\( (1 + \tan A) (1 + \tan B) \)
Multiply the terms:
\( = 1 \cdot 1 + 1 \cdot \tan B + \tan A \cdot 1 + \tan A \cdot \tan B \)
\( = 1 + \tan B + \tan A + \tan A \tan B \)
Rearrange the terms:
\( = 1 + (\tan A + \tan B) + \tan A \tan B \)
From Equation 1, we know that \( (\tan A + \tan B) \) is equal to \( (1 - \tan A \tan B) \). Substitute this into the expression:
\( = 1 + (1 - \tan A \tan B) + \tan A \tan B \)
\( = 1 + 1 - \tan A \tan B + \tan A \tan B \)
The \( \tan A \tan B \) terms cancel each other out:
\( = 2 \)
Hence, \( (1 + \tan A) (1 + \tan B) = 2 \) is proven.
In simple words: We started with the given condition \( A + B = 45^\circ \) and took the tangent of both sides. This helped us find a relationship between \( \tan A \) and \( \tan B \). Then, we expanded the expression \( (1 + \tan A) (1 + \tan B) \) and replaced \( (\tan A + \tan B) \) with what we found earlier. This made the whole expression simplify to 2.

🎯 Exam Tip: This is a very common identity. When you see expressions involving \( (1 + \tan A) \), especially with angles summing to \( 45^\circ \) or \( \pi/4 \), consider using the tangent addition formula as the first step.

Question 7. Prove that \( (1 + \tan 1^\circ) (1 + \tan 2^\circ) (1 + \tan 3^\circ) \dots (1 + \tan 44^\circ) \) is a multiple of 4.
Answer:
From Question 6, we know that if \( A + B = 45^\circ \), then \( (1 + \tan A) (1 + \tan B) = 2 \).
We can pair the terms in the given product:
Pair 1: \( (1 + \tan 1^\circ) (1 + \tan 44^\circ) \)
Since \( 1^\circ + 44^\circ = 45^\circ \), this pair's product is 2.
Pair 2: \( (1 + \tan 2^\circ) (1 + \tan 43^\circ) \)
Since \( 2^\circ + 43^\circ = 45^\circ \), this pair's product is 2.
This pattern continues. The last pair will be:
Last Pair: \( (1 + \tan 22^\circ) (1 + \tan 23^\circ) \)
Since \( 22^\circ + 23^\circ = 45^\circ \), this pair's product is 2.
To find the total number of pairs, we look at the first angle in each pair, which goes from \( 1^\circ \) to \( 22^\circ \). So, there are 22 such pairs.
The entire product is the multiplication of these 22 pairs:
Product \( = (1 + \tan 1^\circ)(1 + \tan 44^\circ) \cdot (1 + \tan 2^\circ)(1 + \tan 43^\circ) \dots (1 + \tan 22^\circ)(1 + \tan 23^\circ) \)
Product \( = 2 \cdot 2 \cdot \dots \cdot 2 \) (22 times)
Product \( = 2^{22} \)
We need to show that this is a multiple of 4.
We can write \( 2^{22} \) as \( 2^2 \cdot 2^{20} \):
Product \( = 4 \cdot 2^{20} \)
Since \( 2^{20} \) is an integer, \( 4 \cdot 2^{20} \) is clearly a multiple of 4.
Thus, the given product is a multiple of 4.
In simple words: We used a special rule that says if two angles add up to 45 degrees, then \( (1 + \tan A) (1 + \tan B) \) equals 2. We found 22 such pairs in the long multiplication. So, the whole multiplication became 2 multiplied by itself 22 times, which is \( 2^{22} \). Since \( 2^{22} \) can be written as \( 4 \times 2^{20} \), it must be a number that can be divided by 4.

🎯 Exam Tip: Recognize patterns in products or sums of trigonometric terms. Often, these problems rely on a specific identity that pairs terms together. This identity, \( (1+\tan A)(1+\tan B)=2 \) when \( A+B=45^\circ \), is a key shortcut.

Question 8. Prove that \( \tan \left(\frac{\pi}{4}+\theta\right) – \tan \left(\frac{\pi}{4}-\theta\right) = 2 \tan 2\theta \)
Answer:
We need to prove the identity \( \tan \left(\frac{\pi}{4}+\theta\right) – \tan \left(\frac{\pi}{4}-\theta\right) = 2 \tan 2\theta \).
Let's start from the left-hand side (LHS).
Using the tangent addition and subtraction formulas:
\( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
\( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
Here, \( A = \frac{\pi}{4} \), so \( \tan A = \tan \frac{\pi}{4} = 1 \).
Therefore, the first term is:
\( \tan \left(\frac{\pi}{4}+\theta\right) = \frac{1 + \tan \theta}{1 - \tan \theta} \)
And the second term is:
\( \tan \left(\frac{\pi}{4}-\theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta} \)
Now, subtract the second term from the first term:
LHS \( = \frac{1 + \tan \theta}{1 - \tan \theta} - \frac{1 - \tan \theta}{1 + \tan \theta} \)
To subtract, find a common denominator, which is \( (1 - \tan \theta)(1 + \tan \theta) \):
LHS \( = \frac{(1 + \tan \theta)^2 - (1 - \tan \theta)^2}{(1 - \tan \theta)(1 + \tan \theta)} \)
Expand the numerator using \( (a+b)^2 - (a-b)^2 = 4ab \), or by direct expansion:
Numerator \( = (1 + 2\tan \theta + \tan^2 \theta) - (1 - 2\tan \theta + \tan^2 \theta) \)
\( = 1 + 2\tan \theta + \tan^2 \theta - 1 + 2\tan \theta - \tan^2 \theta \)
\( = 4\tan \theta \)
Expand the denominator using \( (a-b)(a+b) = a^2 - b^2 \):
Denominator \( = 1^2 - (\tan \theta)^2 = 1 - \tan^2 \theta \)
So, the LHS simplifies to:
LHS \( = \frac{4\tan \theta}{1 - \tan^2 \theta} \)
We know the double angle formula for tangent: \( \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \)
We can rewrite our simplified LHS using this formula:
LHS \( = 2 \cdot \left( \frac{2\tan \theta}{1 - \tan^2 \theta} \right) \)
LHS \( = 2 \tan 2\theta \)
This is equal to the right-hand side (RHS). Thus, the identity is proven.
In simple words: We started by writing out both parts of the left side using the tangent addition and subtraction formulas. Since \( \tan(\pi/4) \) is 1, these expressions became simpler. Then, we subtracted them by finding a common bottom part. The top part simplified to \( 4\tan\theta \) and the bottom part simplified to \( 1-\tan^2\theta \). This whole fraction is equal to \( 2 \tan 2\theta \), which matches the right side of the equation.

🎯 Exam Tip: Always remember the tangent sum and difference formulas. The terms \( \tan(\frac{\pi}{4}+\theta) \) and \( \tan(\frac{\pi}{4}-\theta) \) often simplify very quickly due to \( \tan \frac{\pi}{4} = 1 \).

Question 9. Show that \( \cot \left(7\frac{1}{2}^\circ\right) = \sqrt{2} + \sqrt{3} + \sqrt{8} + \sqrt{6} \)
Answer:
We need to find the value of \( \cot \left(7\frac{1}{2}^\circ\right) \). We can write \( 7\frac{1}{2}^\circ \) as \( \frac{15}{2}^\circ \).
We use the half-angle formula for cotangent:
\( \cot \left(\frac{A}{2}\right) = \frac{\sin A}{1 - \cos A} \)
Let \( A = 15^\circ \). Then \( \frac{A}{2} = 7\frac{1}{2}^\circ \).
First, we need the values of \( \sin 15^\circ \) and \( \cos 15^\circ \).
We know that \( 15^\circ = 45^\circ - 30^\circ \).
\( \sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \)
\( = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
\( \cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \)
\( = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
Now, substitute these values into the cotangent formula:
\( \cot \left(7\frac{1}{2}^\circ\right) = \frac{\sin 15^\circ}{1 - \cos 15^\circ} \)
\( = \frac{\frac{\sqrt{3} - 1}{2\sqrt{2}}}{1 - \frac{\sqrt{3} + 1}{2\sqrt{2}}} \)
Simplify the denominator:
\( = \frac{\frac{\sqrt{3} - 1}{2\sqrt{2}}}{\frac{2\sqrt{2} - (\sqrt{3} + 1)}{2\sqrt{2}}} \)
\( = \frac{\sqrt{3} - 1}{2\sqrt{2} - \sqrt{3} - 1} \)
To rationalize the denominator, multiply the numerator and denominator by the conjugate \( (2\sqrt{2} + \sqrt{3} + 1) \):
\( = \frac{(\sqrt{3} - 1)(2\sqrt{2} + \sqrt{3} + 1)}{((2\sqrt{2}) - (\sqrt{3} + 1))((2\sqrt{2}) + (\sqrt{3} + 1))} \)
Expand the numerator:
\( (\sqrt{3} - 1)(2\sqrt{2} + \sqrt{3} + 1) = \sqrt{3}(2\sqrt{2}) + \sqrt{3}(\sqrt{3}) + \sqrt{3}(1) - 1(2\sqrt{2}) - 1(\sqrt{3}) - 1(1) \)
\( = 2\sqrt{6} + 3 + \sqrt{3} - 2\sqrt{2} - \sqrt{3} - 1 \)
\( = 2\sqrt{6} - 2\sqrt{2} + 2 \)
Expand the denominator using \( (X-Y)(X+Y) = X^2 - Y^2 \), where \( X = 2\sqrt{2} \) and \( Y = \sqrt{3} + 1 \):
\( (2\sqrt{2})^2 - (\sqrt{3} + 1)^2 \)
\( = 8 - ((\sqrt{3})^2 + 2\sqrt{3}(1) + 1^2) \)
\( = 8 - (3 + 2\sqrt{3} + 1) \)
\( = 8 - (4 + 2\sqrt{3}) \)
\( = 4 - 2\sqrt{3} \)
Now, put the numerator and denominator back together:
\( \cot \left(7\frac{1}{2}^\circ\right) = \frac{2\sqrt{6} - 2\sqrt{2} + 2}{4 - 2\sqrt{3}} \)
Divide the numerator and denominator by 2:
\( = \frac{\sqrt{6} - \sqrt{2} + 1}{2 - \sqrt{3}} \)
Rationalize again by multiplying by \( (2 + \sqrt{3}) \):
\( = \frac{(\sqrt{6} - \sqrt{2} + 1)(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} \)
Denominator: \( (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \).
Expand the numerator:
\( (\sqrt{6} - \sqrt{2} + 1)(2 + \sqrt{3}) = 2\sqrt{6} + \sqrt{18} - 2\sqrt{2} - \sqrt{6} + 2 + \sqrt{3} \)
\( = 2\sqrt{6} + 3\sqrt{2} - 2\sqrt{2} - \sqrt{6} + 2 + \sqrt{3} \)
Combine like terms:
\( = (2\sqrt{6} - \sqrt{6}) + (3\sqrt{2} - 2\sqrt{2}) + 2 + \sqrt{3} \)
\( = \sqrt{6} + \sqrt{2} + 2 + \sqrt{3} \)
Rearranging the terms and rewriting 2 as \( \sqrt{4} \) (although the question asks for \( \sqrt{8} \), the correct numerical value is 2, or \( \sqrt{4} \)):
\( = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} \)
This matches the numerical value, but with \( \sqrt{4} \) instead of \( \sqrt{8} \). The question statement has a slight error in the target value `\sqrt{8}`.
If the question meant \( \sqrt{8} \), it would be \( 2\sqrt{2} \), making the target value \( 3\sqrt{2} + \sqrt{3} + \sqrt{6} \), which is numerically different from \( \cot(7.5^\circ) \). The derived answer \( \sqrt{2} + \sqrt{3} + 2 + \sqrt{6} \) is correct.
In simple words: We found the cotangent of \( 7.5^\circ \) by using the half-angle formula, where \( A = 15^\circ \). First, we calculated \( \sin 15^\circ \) and \( \cos 15^\circ \). Then we put these values into the cotangent formula. We had to multiply by special numbers (conjugates) twice to remove the square roots from the bottom part of the fraction. After all the steps, we got \( \sqrt{6} + \sqrt{2} + 2 + \sqrt{3} \). This is the correct numerical value.

🎯 Exam Tip: To evaluate trigonometric functions for angles like \( 7.5^\circ \) or \( 22.5^\circ \), use half-angle formulas with known angles (e.g., \( 15^\circ \) or \( 45^\circ \)). Rationalizing the denominator is a key algebraic step to simplify the final answer.

Question 10. Prove that \( (1 + \sec 2\theta) (1 + \sec 4\theta) \dots (1 + \sec 2^n\theta) = \tan 2^n\theta \cot \theta \)
Answer:
Let \( P = (1 + \sec 2\theta) (1 + \sec 4\theta) \dots (1 + \sec 2^n\theta) \).
First, we use the identity \( 1 + \sec X = 1 + \frac{1}{\cos X} = \frac{\cos X + 1}{\cos X} = \frac{2\cos^2(X/2)}{\cos X} \).
Applying this to each term in the product:
Term 1: \( 1 + \sec 2\theta = \frac{2\cos^2\theta}{\cos 2\theta} \)
Term 2: \( 1 + \sec 4\theta = \frac{2\cos^2(2\theta)}{\cos 4\theta} \)
...
Term n: \( 1 + \sec 2^n\theta = \frac{2\cos^2(2^{n-1}\theta)}{\cos 2^n\theta} \)
Now, multiply all these terms to get P:
\( P = \left( \frac{2\cos^2\theta}{\cos 2\theta} \right) \cdot \left( \frac{2\cos^2(2\theta)}{\cos 4\theta} \right) \cdot \dots \cdot \left( \frac{2\cos^2(2^{n-1}\theta)}{\cos 2^n\theta} \right) \)
Combine all the numerators and denominators:
\( P = \frac{2^n \cdot \cos^2\theta \cdot \cos^2(2\theta) \cdot \cos^2(4\theta) \cdot \dots \cdot \cos^2(2^{n-1}\theta)}{\cos 2\theta \cdot \cos 4\theta \cdot \cos 8\theta \cdot \dots \cdot \cos 2^n\theta} \)
Cancel out one power of each cosine term in the numerator and denominator, from \( \cos 2\theta \) up to \( \cos 2^{n-1}\theta \):
\( P = \frac{2^n \cdot \cos^2\theta \cdot (\cos 2\theta \cdot \cos 4\theta \cdot \dots \cdot \cos 2^{n-1}\theta)}{\cos 2^n\theta \cdot (\cos 2\theta \cdot \cos 4\theta \cdot \dots \cdot \cos 2^{n-1}\theta)} \)
So, this simplifies to:
\( P = \frac{2^n \cos^2\theta}{\cos 2^n\theta} \cdot \frac{\cos 2\theta \cos 4\theta \dots \cos 2^{n-1}\theta}{\cos 2\theta \cos 4\theta \dots \cos 2^{n-1}\theta} \)
The common product cancels out, leaving:
\( P = \frac{2^n \cos^2\theta}{\cos 2^n\theta} \)
Now we need to show that this equals \( \tan 2^n\theta \cot \theta \).
Recall the product identity: \( \cos A \cos 2A \cos 4A \dots \cos 2^{n-1}A = \frac{\sin 2^n A}{2^n \sin A} \).
We can rewrite \( P \) as:
\( P = \frac{2^n \cos\theta}{\cos 2^n\theta} \cdot \cos\theta \)
We want to introduce \( \sin\theta \) in the denominator to form \( \cot\theta \), and \( \sin 2^n\theta \) in the numerator to form \( \tan 2^n\theta \).
Consider the expression: \( \frac{\sin 2^n \theta}{2^n \sin \theta} = \cos \theta \cos 2\theta \cos 4\theta \dots \cos 2^{n-1}\theta \).
So, \( \sin 2^n \theta = 2^n \sin \theta \cos \theta \cos 2\theta \dots \cos 2^{n-1}\theta \).
Let's rewrite \( P \) again:
\( P = \frac{2^n \cos\theta}{\cos 2^n\theta} \cdot \cos\theta \)
Multiply numerator and denominator by \( \sin \theta \):
\( P = \frac{2^n \cos\theta \cdot \sin \theta \cdot \cos\theta}{\cos 2^n\theta \cdot \sin \theta} \)
This path is getting complex. Let's use the product identity directly from where we had \( P = \frac{2^n \cos^2\theta \cos(2\theta) \dots \cos(2^{n-1}\theta)}{\cos 2^n\theta} \).
From the cancellation step, we had:
\( P = \frac{2^n \cos^2\theta \cos(2\theta) \dots \cos(2^{n-1}\theta)}{\cos 2^n\theta} \)
Rearrange this to use the product identity:
\( P = \frac{\cos\theta}{\sin\theta} \cdot \frac{2^n \sin\theta \cos\theta \cos(2\theta) \dots \cos(2^{n-1}\theta)}{\cos 2^n\theta} \)
Recognize the numerator part \( 2^n \sin\theta \cos\theta \cos(2\theta) \dots \cos(2^{n-1}\theta) \) as the expansion of \( \sin(2^n\theta) \).
Thus, \( P = \frac{\cos\theta}{\sin\theta} \cdot \frac{\sin(2^n\theta)}{\cos(2^n\theta)} \)
\( P = \cot \theta \cdot \tan 2^n\theta \)
This proves the desired result.
In simple words: We took each term like \( (1 + \sec X) \) and changed it using a specific trigonometric identity. When we multiplied all these changed terms, many parts cancelled out. Then, we used a known identity for the product of cosine terms to simplify the remaining expression, which eventually gave us \( \tan 2^n\theta \cot \theta \).

🎯 Exam Tip: Problems involving products of trigonometric functions with powers of 2 in the angle often use the identity \( \cos A \cos 2A \dots \cos 2^{n-1}A = \frac{\sin 2^n A}{2^n \sin A} \). Look for opportunities to apply this identity.

Question 11. Prove that \( 32\sqrt{3} \sin \frac{\pi}{48} \cos \frac{\pi}{48} \cos \frac{\pi}{24} \cos \frac{\pi}{12} \cos \frac{\pi}{6} = 3 \)
Answer:
We need to prove that the given expression equals 3.
Let's start from the left-hand side (LHS) and use the double angle identity \( 2 \sin X \cos X = \sin 2X \) repeatedly.
LHS \( = 32\sqrt{3} \left(\sin \frac{\pi}{48} \cos \frac{\pi}{48}\right) \cos \frac{\pi}{24} \cos \frac{\pi}{12} \cos \frac{\pi}{6} \)
We can rewrite the first pair using the identity \( 2 \sin X \cos X = \sin 2X \). For this, we need to extract a 2 from 32.
LHS \( = 16\sqrt{3} \left(2 \sin \frac{\pi}{48} \cos \frac{\pi}{48}\right) \cos \frac{\pi}{24} \cos \frac{\pi}{12} \cos \frac{\pi}{6} \)
Apply the identity:
LHS \( = 16\sqrt{3} \sin \left(2 \cdot \frac{\pi}{48}\right) \cos \frac{\pi}{24} \cos \frac{\pi}{12} \cos \frac{\pi}{6} \)
LHS \( = 16\sqrt{3} \sin \frac{\pi}{24} \cos \frac{\pi}{24} \cos \frac{\pi}{12} \cos \frac{\pi}{6} \)
Repeat the process with the next pair \( \sin \frac{\pi}{24} \cos \frac{\pi}{24} \). Extract another 2 from 16:
LHS \( = 8\sqrt{3} \left(2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}\right) \cos \frac{\pi}{12} \cos \frac{\pi}{6} \)
Apply the identity again:
LHS \( = 8\sqrt{3} \sin \left(2 \cdot \frac{\pi}{24}\right) \cos \frac{\pi}{12} \cos \frac{\pi}{6} \)
LHS \( = 8\sqrt{3} \sin \frac{\pi}{12} \cos \frac{\pi}{12} \cos \frac{\pi}{6} \)
Repeat with \( \sin \frac{\pi}{12} \cos \frac{\pi}{12} \). Extract another 2 from 8:
LHS \( = 4\sqrt{3} \left(2 \sin \frac{\pi}{12} \cos \frac{\pi}{12}\right) \cos \frac{\pi}{6} \)
Apply the identity:
LHS \( = 4\sqrt{3} \sin \left(2 \cdot \frac{\pi}{12}\right) \cos \frac{\pi}{6} \)
LHS \( = 4\sqrt{3} \sin \frac{\pi}{6} \cos \frac{\pi}{6} \)
Repeat with \( \sin \frac{\pi}{6} \cos \frac{\pi}{6} \). Extract another 2 from 4:
LHS \( = 2\sqrt{3} \left(2 \sin \frac{\pi}{6} \cos \frac{\pi}{6}\right) \)
Apply the identity:
LHS \( = 2\sqrt{3} \sin \left(2 \cdot \frac{\pi}{6}\right) \)
LHS \( = 2\sqrt{3} \sin \frac{\pi}{3} \)
Now, substitute the known value of \( \sin \frac{\pi}{3} \) (which is \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)):
LHS \( = 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} \)
LHS \( = \sqrt{3} \cdot \sqrt{3} \)
LHS \( = 3 \)
This is equal to the right-hand side. Thus, the identity is proven.
In simple words: We used the double angle formula \( 2 \sin X \cos X = \sin 2X \) multiple times. We kept pairing up \( \sin \) and \( \cos \) terms and changing them into a single \( \sin \) term with a doubled angle. We started with \( \sin \frac{\pi}{48} \cos \frac{\pi}{48} \), then moved to \( \sin \frac{\pi}{24} \cos \frac{\pi}{24} \), and so on, until we reached \( \sin \frac{\pi}{3} \). Finally, we put the value of \( \sin \frac{\pi}{3} \) to get the answer 3.

🎯 Exam Tip: For products of sine and cosine with angles in geometric progression (like \( X, 2X, 4X, \dots \)), always think of applying the double angle identity \( 2 \sin A \cos A = \sin 2A \) repeatedly. Make sure to keep track of the coefficients (like 32, 16, 8, etc.).