OP Malhotra Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Chapter Test

S Chand Class 11 ICSE Maths Solutions Chapter 5 Compound and Multiple Angles Chapter Test

 

Question 1. Show that tan 75° = \( \frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3} \). Hence deduce that tan 75° – cot 75° = 4 sin 60°.
Answer: We begin by using the tangent addition formula. We know that \( \tan 75^\circ = \tan (45^\circ + 30^\circ) \).
\( \implies \tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
So, \( \tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \)
We substitute the known values: \( \tan 45^\circ = 1 \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
\( \implies \tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \)
To simplify this expression, we multiply the numerator and denominator by the conjugate of the denominator, \( \sqrt{3}+1 \).
\( \implies \tan 75^\circ = \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{(\sqrt{3})^2 + 1^2 + 2\sqrt{3}}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \)
Now, we need to deduce the value of \( \tan 75^\circ - \cot 75^\circ \).
We know that \( \cot 75^\circ = \frac{1}{\tan 75^\circ} = \frac{1}{2+\sqrt{3}} \).
To simplify \( \frac{1}{2+\sqrt{3}} \), we multiply the numerator and denominator by its conjugate, \( 2-\sqrt{3} \).
\( \implies \cot 75^\circ = \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2-\sqrt{3}}{4-3} = 2-\sqrt{3} \)
So, \( \tan 75^\circ - \cot 75^\circ = (2+\sqrt{3}) - (2-\sqrt{3}) = 2+\sqrt{3}-2+\sqrt{3} = 2\sqrt{3} \).
The final step is to show that \( 2\sqrt{3} = 4 \sin 60^\circ \).
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
So, \( 4 \sin 60^\circ = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \).
Therefore, \( \tan 75^\circ - \cot 75^\circ = 4 \sin 60^\circ \). Both sides simplify to \( 2\sqrt{3} \). This shows the identity holds true.
In simple words: First, we find the value of tan 75 degrees using a formula for adding angles. We then simplify this value. Next, we find cot 75 degrees and subtract it from tan 75 degrees. Finally, we check if this answer is the same as 4 times sin 60 degrees.

🎯 Exam Tip: When proving trigonometric identities, start with the more complex side and simplify it. Remember to rationalize denominators and use known trigonometric values for standard angles.

 

Question 2. Prove that sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2)x = cos x.
Answer: We need to prove the given trigonometric identity. Let's start with the left-hand side (L.H.S.).
L.H.S. \( = \sin (n+1)x \sin (n+2)x + \cos (n+1)x \cos (n+2)x \)
This expression looks like the cosine difference formula, which is \( \cos (A-B) = \cos A \cos B + \sin A \sin B \).
Here, we can let \( A = (n+2)x \) and \( B = (n+1)x \).
Then, the L.H.S. can be written as \( \cos [(n+2)x - (n+1)x] \).
Now, we simplify the expression inside the cosine:
\( (n+2)x - (n+1)x = nx + 2x - nx - x = x \).
So, L.H.S. \( = \cos x \).
This is equal to the right-hand side (R.H.S.).
Thus, the identity \( \sin (n+1)x \sin (n+2)x + \cos (n+1)x \cos (n+2)x = \cos x \) is proven. This formula is very useful for simplifying expressions involving products of sines and cosines.
In simple words: We start with the left side of the equation. We notice it matches a known formula for cosine of the difference of two angles. When we apply that formula and simplify, we get 'cos x', which is the right side of the equation.

🎯 Exam Tip: Recognize standard trigonometric formulas like the sum/difference identities for sine and cosine. This question is a direct application of the \( \cos(A-B) \) formula.

 

Question 3. If A + B + C = π, and cos A = cos B cos C, show that 2 cot B cot C = 1.
Answer: We are given two conditions: \( A + B + C = \pi \) and \( \cos A = \cos B \cos C \). We need to show that \( 2 \cot B \cot C = 1 \).
From the first condition, \( A + B + C = \pi \), we can write \( A = \pi - (B+C) \).
Now substitute this into the second condition: \( \cos A = \cos B \cos C \).
\( \implies \cos [\pi - (B+C)] = \cos B \cos C \)
We know that \( \cos (\pi - \theta) = - \cos \theta \).
So, \( - \cos (B+C) = \cos B \cos C \).
Next, expand \( \cos (B+C) \) using the sum formula: \( \cos (B+C) = \cos B \cos C - \sin B \sin C \).
\( \implies - (\cos B \cos C - \sin B \sin C) = \cos B \cos C \)
\( \implies - \cos B \cos C + \sin B \sin C = \cos B \cos C \)
Now, rearrange the terms to gather similar trigonometric functions:
\( \implies \sin B \sin C = \cos B \cos C + \cos B \cos C \)
\( \implies \sin B \sin C = 2 \cos B \cos C \)
To get cotangent terms, we divide both sides by \( \sin B \sin C \).
\( \implies \frac{\sin B \sin C}{\sin B \sin C} = \frac{2 \cos B \cos C}{\sin B \sin C} \)
\( \implies 1 = 2 \left( \frac{\cos B}{\sin B} \right) \left( \frac{\cos C}{\sin C} \right) \)
Since \( \frac{\cos \theta}{\sin \theta} = \cot \theta \), we have:
\( \implies 1 = 2 \cot B \cot C \)
Thus, we have successfully shown that \( 2 \cot B \cot C = 1 \). This demonstrates how angle relationships in a triangle can lead to specific trigonometric identities.
In simple words: We start by using the fact that the angles A, B, and C add up to 180 degrees. This helps us change 'cos A' into something related to 'B' and 'C'. Then, we use the given condition and basic trigonometry to rearrange the terms until we get '2 cot B cot C = 1'.

🎯 Exam Tip: When given a condition like \( A+B+C=\pi \), immediately think of substitutions like \( A = \pi - (B+C) \) or \( B+C = \pi - A \) to simplify the expression using angle properties of a triangle.

 

Question 4. Show that \( \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)} = 3 \), given that \( \tan \alpha = 2 \tan \beta \).
Answer: We are given that \( \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)} = 3 \) needs to be proven, with the condition \( \tan \alpha = 2 \tan \beta \).
Let's start with the given equation \( \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)} = 3 \). We can write this as:
\( \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)} = \frac{3}{1} \)
Now, we apply the componendo and dividendo rule on both sides. This rule states that if \( \frac{a}{b} = \frac{c}{d} \), then \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \).
Applying this rule, we get:
\( \frac{\sin (\alpha+\beta) + \sin (\alpha-\beta)}{\sin (\alpha+\beta) - \sin (\alpha-\beta)} = \frac{3+1}{3-1} \)
\( \implies \frac{\sin (\alpha+\beta) + \sin (\alpha-\beta)}{\sin (\alpha+\beta) - \sin (\alpha-\beta)} = \frac{4}{2} = 2 \)
Now, we use the sum and difference formulas for sine:
\( \sin (A+B) + \sin (A-B) = 2 \sin A \cos B \)
\( \sin (A+B) - \sin (A-B) = 2 \cos A \sin B \)
Applying these to our equation (with \( A=\alpha \) and \( B=\beta \)):
\( \frac{2 \sin \alpha \cos \beta}{2 \cos \alpha \sin \beta} = 2 \)
The \( 2 \) in the numerator and denominator cancel out:
\( \implies \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta} = 2 \)
We can rewrite this using the definition of tangent and cotangent:
\( \implies \left( \frac{\sin \alpha}{\cos \alpha} \right) \left( \frac{\cos \beta}{\sin \beta} \right) = 2 \)
\( \implies \tan \alpha \cot \beta = 2 \)
Since \( \cot \beta = \frac{1}{\tan \beta} \), we have:
\( \implies \frac{\tan \alpha}{\tan \beta} = 2 \)
\( \implies \tan \alpha = 2 \tan \beta \)
This matches the given condition. Since applying componendo and dividendo and simplifying leads back to the given condition, the initial statement must be true. This shows how useful algebraic tools like componendo and dividendo are in trigonometry.
In simple words: We start with the fraction involving sine of sum and difference of angles. We use a math rule called componendo and dividendo. Then we simplify the expression using sine addition and subtraction formulas. This leads us to a simple ratio of tan alpha and tan beta, which matches what was given in the problem, proving the first statement.

🎯 Exam Tip: When faced with ratios of trigonometric functions, consider using componendo and dividendo, as it often simplifies expressions into forms involving products or ratios of single functions.

 

Question 5. Show that \( \frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}} = \tan 55^{\circ} \).
Answer: We need to show that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}} \)
To simplify this expression and get it into a tangent form, we divide both the numerator and the denominator by \( \cos 10^{\circ} \).
\( \implies \frac{\frac{\cos 10^{\circ}}{\cos 10^{\circ}} + \frac{\sin 10^{\circ}}{\cos 10^{\circ}}}{\frac{\cos 10^{\circ}}{\cos 10^{\circ}} - \frac{\sin 10^{\circ}}{\cos 10^{\circ}}} \)
\( \implies \frac{1 + \tan 10^{\circ}}{1 - \tan 10^{\circ}} \)
We know that \( \tan 45^{\circ} = 1 \). So, we can replace the '1' in the numerator with \( \tan 45^{\circ} \). We also know that \( 1 \cdot \tan 10^\circ = \tan 10^\circ \), so we can multiply the \( \tan 10^\circ \) in the denominator by \( \tan 45^\circ \).
\( \implies \frac{\tan 45^{\circ} + \tan 10^{\circ}}{1 - \tan 45^{\circ} \tan 10^{\circ}} \)
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 \( B = 10^{\circ} \).
\( \implies \tan (45^{\circ} + 10^{\circ}) \)
\( \implies \tan 55^{\circ} \)
This is equal to the R.H.S.
Thus, the identity \( \frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}} = \tan 55^{\circ} \) is proven. This shows a common trick in trigonometry to convert sums/differences of sines and cosines into tangents.
In simple words: We take the left side of the equation and divide every term by cos 10 degrees. This changes the expression into a form that looks like the formula for 'tan (A+B)'. By recognizing that 1 is equal to tan 45 degrees, we can then rewrite the expression as tan (45 + 10) degrees, which is tan 55 degrees, matching the right side.

🎯 Exam Tip: A common technique to simplify expressions like \( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \) is to divide both numerator and denominator by \( \cos \theta \), transforming the expression into terms of \( \tan \theta \).

 

Question 6. If \( \sin 2A = \frac { 4 }{ 5 } \), find the value of tan A, \( (0^{\circ} \le A \le \frac { \pi }{ 4 }) \)
Answer: We are given \( \sin 2A = \frac{4}{5} \) and that \( A \) is an angle between \( 0^\circ \) and \( \frac{\pi}{4} \) (which is \( 45^\circ \)). We need to find the value of \( \tan A \).
We use the double angle formula for sine in terms of tangent: \( \sin 2A = \frac{2 \tan A}{1 + \tan^2 A} \).
Substitute the given value:
\( \frac{4}{5} = \frac{2 \tan A}{1 + \tan^2 A} \)
Now, cross-multiply:
\( 4(1 + \tan^2 A) = 5(2 \tan A) \)
\( 4 + 4 \tan^2 A = 10 \tan A \)
Rearrange this into a quadratic equation by moving all terms to one side:
\( 4 \tan^2 A - 10 \tan A + 4 = 0 \)
We can simplify this equation by dividing all terms by 2:
\( 2 \tan^2 A - 5 \tan A + 2 = 0 \)
Let \( x = \tan A \). The equation becomes \( 2x^2 - 5x + 2 = 0 \).
We can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a=2 \), \( b=-5 \), \( c=2 \).
\( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} \)
\( x = \frac{5 \pm \sqrt{25 - 16}}{4} \)
\( x = \frac{5 \pm \sqrt{9}}{4} \)
\( x = \frac{5 \pm 3}{4} \)
This gives two possible values for \( \tan A \):
\( \tan A = \frac{5+3}{4} = \frac{8}{4} = 2 \)
or
\( \tan A = \frac{5-3}{4} = \frac{2}{4} = \frac{1}{2} \)
Now, we use the given condition \( 0^\circ \le A \le \frac{\pi}{4} \) or \( 0^\circ \le A \le 45^\circ \).
In the first quadrant, the tangent function is increasing.
If \( A = 45^\circ \), then \( \tan A = \tan 45^\circ = 1 \).
Since \( A \le 45^\circ \), \( \tan A \) must be less than or equal to 1.
Therefore, \( \tan A = 2 \) is not a valid solution because \( 2 > 1 \).
The only valid solution is \( \tan A = \frac{1}{2} \). This calculation shows how crucial the given range of the angle is for finding the correct solution.
In simple words: We use a special formula that connects sin 2A with tan A. We plug in the given value for sin 2A and solve the equation for tan A. We get two possible answers. Because angle A is between 0 and 45 degrees, tan A must be less than or equal to 1. This helps us pick the correct value for tan A from the two answers.

🎯 Exam Tip: Always consider the given range of the angle carefully, especially when solving trigonometric equations. It helps eliminate extraneous solutions that arise from the algebraic manipulation.

 

Question 7. Express (i) cot A in terms of cos 2 A, (ii) cos 4θ in terms of cos θ.
Answer:
(i) To express \( \cot A \) in terms of \( \cos 2A \):
We know the identity \( \cot^2 A = \frac{\cos^2 A}{\sin^2 A} \).
We also know the half-angle identities for \( \cos^2 A \) and \( \sin^2 A \) in terms of \( \cos 2A \):
\( \cos^2 A = \frac{1 + \cos 2A}{2} \)
\( \sin^2 A = \frac{1 - \cos 2A}{2} \)
Now, substitute these into the expression for \( \cot^2 A \):
\( \cot^2 A = \frac{\frac{1 + \cos 2A}{2}}{\frac{1 - \cos 2A}{2}} \)
The 2's in the denominators cancel out:
\( \cot^2 A = \frac{1 + \cos 2A}{1 - \cos 2A} \)
To find \( \cot A \), we take the square root of both sides:
\( \cot A = \pm \sqrt{\frac{1 + \cos 2A}{1 - \cos 2A}} \)
The sign (\( \pm \)) depends on the quadrant of angle A. This formula connects cotangent of an angle to the cosine of its double angle.
(ii) To express \( \cos 4\theta \) in terms of \( \cos \theta \):
We will use the double angle formula for cosine: \( \cos 2x = 2 \cos^2 x - 1 \).
Let \( x = 2\theta \). Then \( \cos 4\theta = \cos (2 \times 2\theta) = 2 \cos^2 (2\theta) - 1 \).
Now, we need to express \( \cos (2\theta) \) in terms of \( \cos \theta \).
Again, use the double angle formula: \( \cos 2\theta = 2 \cos^2 \theta - 1 \).
Substitute this expression for \( \cos 2\theta \) back into the equation for \( \cos 4\theta \):
\( \cos 4\theta = 2 (2 \cos^2 \theta - 1)^2 - 1 \)
Expand the squared term \( (2 \cos^2 \theta - 1)^2 \):
\( (2 \cos^2 \theta - 1)^2 = (2 \cos^2 \theta)^2 - 2(2 \cos^2 \theta)(1) + 1^2 = 4 \cos^4 \theta - 4 \cos^2 \theta + 1 \)
Now, substitute this back:
\( \cos 4\theta = 2 (4 \cos^4 \theta - 4 \cos^2 \theta + 1) - 1 \)
Distribute the 2:
\( \cos 4\theta = 8 \cos^4 \theta - 8 \cos^2 \theta + 2 - 1 \)
\( \cos 4\theta = 8 \cos^4 \theta - 8 \cos^2 \theta + 1 \)
This is the expression for \( \cos 4\theta \) in terms of \( \cos \theta \). This formula demonstrates how to repeatedly apply double-angle identities.
In simple words: (i) We write cot A using cos squared and sin squared. Then, we replace cos squared A and sin squared A with formulas that include cos 2A. This gives us cot A using only cos 2A. (ii) We find cos 4θ by first writing it as cos (2 times 2θ). Then we use the formula for cos 2θ twice, each time replacing the terms with those involving cos θ, until the final answer only has cos θ.

🎯 Exam Tip: Remember the fundamental half-angle and double-angle identities. For higher multiples (like \( \cos 4\theta \)), apply the double-angle formula iteratively, step-by-step.

 

Question 8. A positive acute angle is divided into two parts whose tangents are \( \frac {1}{2 } \) and \( \frac { 1 }{ 3 } \). Show that the angle is \( \frac { \pi }{ 4 } \).
Answer: Let the positive acute angle be \( \theta \). We are told that this angle is divided into two parts, let's call them \( \alpha \) and \( \beta \).
So, \( \theta = \alpha + \beta \).
We are given that the tangents of these two parts are \( \frac{1}{2} \) and \( \frac{1}{3} \).
Let \( \tan \alpha = \frac{1}{2} \) and \( \tan \beta = \frac{1}{3} \).
To find the tangent of the sum of these two angles, \( \tan \theta = \tan (\alpha + \beta) \), we use the tangent addition formula:
\( \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \)
Substitute the given values of \( \tan \alpha \) and \( \tan \beta \):
\( \tan \theta = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} \)
First, calculate the sum in the numerator:
\( \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \)
Next, calculate the product in the denominator:
\( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \)
Now, substitute these back into the formula:
\( \tan \theta = \frac{\frac{5}{6}}{1 - \frac{1}{6}} \)
Calculate the denominator:
\( 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \)
So, \( \tan \theta = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \)
We have \( \tan \theta = 1 \).
Since \( \theta \) is a positive acute angle (meaning \( 0^\circ < \theta < 90^\circ \)), the angle whose tangent is 1 is \( 45^\circ \) or \( \frac{\pi}{4} \) radians.
Therefore, \( \theta = \frac{\pi}{4} \). This calculation shows how angle addition formulas can be used to find unknown angles.
In simple words: We are given that an angle is split into two parts, and we know the 'tan' value for each part. We use a formula to find the 'tan' of the total angle by adding these two parts. After putting the numbers into the formula and simplifying, we find that the 'tan' of the total angle is 1. Since it's a sharp angle, this means the angle is 45 degrees, which is the same as pi over 4.

🎯 Exam Tip: When dealing with an angle divided into parts, the tangent addition formula \( \tan(A+B) \) is usually the most direct path to finding the total angle, especially if the tangents of the individual parts are given.

 

Question 9. Show that \( \cos 10^{\circ} + \cos 110^{\circ} + \cos 130^{\circ} = 0 \).
Answer: We need to show that the left-hand side (L.H.S.) equals 0.
L.H.S. \( = \cos 10^{\circ} + \cos 110^{\circ} + \cos 130^{\circ} \)
Let's apply the sum-to-product formula for \( \cos C + \cos D = 2 \cos \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right) \) to the last two terms.
Let \( C = 110^{\circ} \) and \( D = 130^{\circ} \).
\( \frac{C+D}{2} = \frac{110^{\circ}+130^{\circ}}{2} = \frac{240^{\circ}}{2} = 120^{\circ} \)
\( \frac{C-D}{2} = \frac{110^{\circ}-130^{\circ}}{2} = \frac{-20^{\circ}}{2} = -10^{\circ} \)
So, \( \cos 110^{\circ} + \cos 130^{\circ} = 2 \cos 120^{\circ} \cos (-10^{\circ}) \).
We know that \( \cos (-\theta) = \cos \theta \), so \( \cos (-10^{\circ}) = \cos 10^{\circ} \).
Also, \( \cos 120^{\circ} = \cos (180^{\circ} - 60^{\circ}) = - \cos 60^{\circ} = - \frac{1}{2} \).
Substitute these values back:
\( \cos 110^{\circ} + \cos 130^{\circ} = 2 \left( - \frac{1}{2} \right) \cos 10^{\circ} = - \cos 10^{\circ} \).
Now, substitute this back into the original L.H.S. expression:
L.H.S. \( = \cos 10^{\circ} + (-\cos 10^{\circ}) \)
L.H.S. \( = \cos 10^{\circ} - \cos 10^{\circ} = 0 \).
This is equal to the R.H.S.
Thus, the identity is proven. This shows a clever application of trigonometric sum-to-product formulas and angle properties.
In simple words: We start with the three cosine terms. We group the last two terms and use a formula to change their sum into a product. Then we replace the angles with their known values and simplify. We find that the sum of the last two terms exactly cancels out the first term, making the whole expression equal to zero.

🎯 Exam Tip: When you have sums of three or more cosine or sine terms, try to combine two of them using sum-to-product formulas. Look for angles that might simplify nicely or cancel out with other terms, often related to 180 degrees.

 

Question 10. Show that \( \frac{\sin 5 A+2 \sin 8 A+\sin 11 A}{\sin 8 A+2 \sin 11 A+\sin 14 A}=\frac{\sin 8 A}{\sin 11 A} \).
Answer: We need to show that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
L.H.S. \( = \frac{\sin 5 A+2 \sin 8 A+\sin 11 A}{\sin 8 A+2 \sin 11 A+\sin 14 A} \)
Let's rearrange the terms in the numerator and denominator to apply the sum-to-product formula \( \sin C + \sin D = 2 \sin \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right) \).
Numerator: \( (\sin 11A + \sin 5A) + 2 \sin 8A \)
Applying the sum-to-product formula to \( \sin 11A + \sin 5A \):
\( 2 \sin \left( \frac{11A+5A}{2} \right) \cos \left( \frac{11A-5A}{2} \right) = 2 \sin \left( \frac{16A}{2} \right) \cos \left( \frac{6A}{2} \right) = 2 \sin 8A \cos 3A \)
So, the numerator becomes \( 2 \sin 8A \cos 3A + 2 \sin 8A \).
Factor out \( 2 \sin 8A \): \( 2 \sin 8A (\cos 3A + 1) \).
Denominator: \( (\sin 14A + \sin 8A) + 2 \sin 11A \)
Applying the sum-to-product formula to \( \sin 14A + \sin 8A \):
\( 2 \sin \left( \frac{14A+8A}{2} \right) \cos \left( \frac{14A-8A}{2} \right) = 2 \sin \left( \frac{22A}{2} \right) \cos \left( \frac{6A}{2} \right) = 2 \sin 11A \cos 3A \)
So, the denominator becomes \( 2 \sin 11A \cos 3A + 2 \sin 11A \).
Factor out \( 2 \sin 11A \): \( 2 \sin 11A (\cos 3A + 1) \).
Now, substitute these back into the L.H.S. expression:
L.H.S. \( = \frac{2 \sin 8A (\cos 3A + 1)}{2 \sin 11A (\cos 3A + 1)} \)
Assuming \( \cos 3A + 1 \neq 0 \), we can cancel out the common terms \( 2 \) and \( (\cos 3A + 1) \).
L.H.S. \( = \frac{\sin 8A}{\sin 11A} \)
This is equal to the R.H.S.
Thus, the identity is proven. This illustrates the power of factoring and sum-to-product identities in simplifying complex trigonometric fractions.
In simple words: We take the left side of the equation. In the top part (numerator), we group two terms and use a formula to change their sum into a product, then add the remaining term. We do the same for the bottom part (denominator). After factoring out common parts in both the top and bottom, many terms cancel each other out, leaving us with the right side of the equation.

🎯 Exam Tip: When a fraction has sums of trigonometric terms, look for opportunities to apply sum-to-product or product-to-sum formulas. Often, terms will factor out or cancel, simplifying the expression significantly.

 

Question 11. Show that \( \frac{1}{2 \sin 10^{\circ}} – 2 \sin 70^{\circ} = 1 \).
Answer: We need to show that the left-hand side (L.H.S.) equals 1.
L.H.S. \( = \frac{1}{2 \sin 10^{\circ}} - 2 \sin 70^{\circ} \)
To combine these terms, find a common denominator:
L.H.S. \( = \frac{1 - (2 \sin 70^{\circ})(2 \sin 10^{\circ})}{2 \sin 10^{\circ}} \)
L.H.S. \( = \frac{1 - 2 (2 \sin 70^{\circ} \sin 10^{\circ})}{2 \sin 10^{\circ}} \)
Now, use the product-to-sum formula: \( 2 \sin A \sin B = \cos (A-B) - \cos (A+B) \).
Here, \( A = 70^{\circ} \) and \( B = 10^{\circ} \).
\( A-B = 70^{\circ}-10^{\circ} = 60^{\circ} \)
\( A+B = 70^{\circ}+10^{\circ} = 80^{\circ} \)
So, \( 2 \sin 70^{\circ} \sin 10^{\circ} = \cos 60^{\circ} - \cos 80^{\circ} \).
Substitute this back into the L.H.S. expression:
L.H.S. \( = \frac{1 - 2 (\cos 60^{\circ} - \cos 80^{\circ})}{2 \sin 10^{\circ}} \)
We know that \( \cos 60^{\circ} = \frac{1}{2} \).
L.H.S. \( = \frac{1 - 2 \left( \frac{1}{2} - \cos 80^{\circ} \right)}{2 \sin 10^{\circ}} \)
Distribute the 2 in the numerator:
L.H.S. \( = \frac{1 - (2 \times \frac{1}{2}) + (2 \cos 80^{\circ})}{2 \sin 10^{\circ}} \)
L.H.S. \( = \frac{1 - 1 + 2 \cos 80^{\circ}}{2 \sin 10^{\circ}} \)
L.H.S. \( = \frac{2 \cos 80^{\circ}}{2 \sin 10^{\circ}} \)
The 2's cancel out:
L.H.S. \( = \frac{\cos 80^{\circ}}{\sin 10^{\circ}} \)
We know that \( \cos (90^{\circ} - \theta) = \sin \theta \).
So, \( \cos 80^{\circ} = \cos (90^{\circ} - 10^{\circ}) = \sin 10^{\circ} \).
Substitute this into the expression:
L.H.S. \( = \frac{\sin 10^{\circ}}{\sin 10^{\circ}} = 1 \)
This is equal to the R.H.S.
Thus, the identity is proven. This example demonstrates how to use product-to-sum identities and co-function relationships to simplify expressions.
In simple words: We start with the left side of the equation and combine the two terms by finding a common bottom part. Then we use a special formula to change the product of two sine terms into a difference of cosine terms. We use known values for cosines and simplify the expression. Finally, we change the cosine in the numerator to a sine, which then cancels out the sine in the denominator, leaving us with 1.

🎯 Exam Tip: When you see products of sine or cosine terms, think about using product-to-sum formulas. Also, look for complementary angles (summing to 90 degrees) where \( \sin \theta = \cos (90^\circ - \theta) \) can simplify the expression.

 

Question 12. Show that \( \sin 19^{\circ} + \sin 41^{\circ} + \sin 83^{\circ} = \sin 23^{\circ} + \sin 37^{\circ} + \sin 79^{\circ} \).
Answer: We need to show that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
Let's simplify the L.H.S.: \( \sin 19^{\circ} + \sin 41^{\circ} + \sin 83^{\circ} \).
Apply the sum-to-product formula \( \sin C + \sin D = 2 \sin \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right) \) to \( \sin 19^{\circ} + \sin 41^{\circ} \).
\( C = 41^{\circ}, D = 19^{\circ} \)
\( \frac{C+D}{2} = \frac{41^{\circ}+19^{\circ}}{2} = \frac{60^{\circ}}{2} = 30^{\circ} \)
\( \frac{C-D}{2} = \frac{41^{\circ}-19^{\circ}}{2} = \frac{22^{\circ}}{2} = 11^{\circ} \)
So, \( \sin 19^{\circ} + \sin 41^{\circ} = 2 \sin 30^{\circ} \cos 11^{\circ} \).
We know \( \sin 30^{\circ} = \frac{1}{2} \).
\( 2 \sin 30^{\circ} \cos 11^{\circ} = 2 \left( \frac{1}{2} \right) \cos 11^{\circ} = \cos 11^{\circ} \).
Therefore, L.H.S. \( = \cos 11^{\circ} + \sin 83^{\circ} \).
We can write \( \cos 11^{\circ} \) as \( \sin (90^{\circ} - 11^{\circ}) = \sin 79^{\circ} \).
So, L.H.S. \( = \sin 79^{\circ} + \sin 83^{\circ} \). (Equation 1)
Now, let's simplify the R.H.S.: \( \sin 23^{\circ} + \sin 37^{\circ} + \sin 79^{\circ} \).
Apply the sum-to-product formula to \( \sin 23^{\circ} + \sin 37^{\circ} \).
\( C = 37^{\circ}, D = 23^{\circ} \)
\( \frac{C+D}{2} = \frac{37^{\circ}+23^{\circ}}{2} = \frac{60^{\circ}}{2} = 30^{\circ} \)
\( \frac{C-D}{2} = \frac{37^{\circ}-23^{\circ}}{2} = \frac{14^{\circ}}{2} = 7^{\circ} \)
So, \( \sin 23^{\circ} + \sin 37^{\circ} = 2 \sin 30^{\circ} \cos 7^{\circ} \).
Again, \( \sin 30^{\circ} = \frac{1}{2} \).
\( 2 \sin 30^{\circ} \cos 7^{\circ} = 2 \left( \frac{1}{2} \right) \cos 7^{\circ} = \cos 7^{\circ} \).
Therefore, R.H.S. \( = \cos 7^{\circ} + \sin 79^{\circ} \).
We can write \( \cos 7^{\circ} \) as \( \sin (90^{\circ} - 7^{\circ}) = \sin 83^{\circ} \).
So, R.H.S. \( = \sin 83^{\circ} + \sin 79^{\circ} \). (Equation 2)
From (Equation 1) and (Equation 2), we see that L.H.S. \( = \sin 79^{\circ} + \sin 83^{\circ} \) and R.H.S. \( = \sin 83^{\circ} + \sin 79^{\circ} \).
Since L.H.S. \( = \) R.H.S., the identity is proven. This detailed step-by-step simplification highlights the utility of sum-to-product formulas for multiple terms.
In simple words: We work on both sides of the equation separately. For the left side, we combine two sine terms using a sum-to-product formula, then convert the remaining cosine term to sine. We do the same for the right side. After simplifying both sides, we find they become identical, proving the statement.

🎯 Exam Tip: When proving identities with sums of three terms, focus on combining two terms with a sum-to-product formula first. Then, look for ways to simplify the remaining terms using co-function identities or other angle properties.

 

Question 13. If \( \sin A = \frac{1}{\sqrt{3}} \) and \( \sin B = \frac{1}{\sqrt{5}} \) find the value of \( \tan \frac { 1 }{ 2 }(A + B)\cot \frac { 1 }{ 2 }(A – B) \).
Answer: We are given \( \sin A = \frac{1}{\sqrt{3}} \) and \( \sin B = \frac{1}{\sqrt{5}} \). We need to find the value of \( \tan \frac{1}{2}(A+B) \cot \frac{1}{2}(A-B) \).
Let's use the identity:
\( \tan \frac{A+B}{2} \cot \frac{A-B}{2} = \frac{\sin \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)} \cdot \frac{\cos \left(\frac{A-B}{2}\right)}{\sin \left(\frac{A-B}{2}\right)} \)
This expression is equivalent to \( \frac{\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)} \).
Now, we use the product-to-sum formulas:
Numerator: \( 2 \sin X \cos Y = \sin (X+Y) + \sin (X-Y) \)
Denominator: \( 2 \cos X \sin Y = \sin (X+Y) - \sin (X-Y) \)
Here, \( X = \frac{A+B}{2} \) and \( Y = \frac{A-B}{2} \).
\( X+Y = \frac{A+B}{2} + \frac{A-B}{2} = \frac{A+B+A-B}{2} = \frac{2A}{2} = A \)
\( X-Y = \frac{A+B}{2} - \frac{A-B}{2} = \frac{A+B-A+B}{2} = \frac{2B}{2} = B \)
So, \( \tan \frac{A+B}{2} \cot \frac{A-B}{2} = \frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)} = \frac{\sin A + \sin B}{\sin A - \sin B} \)
Now, substitute the given values \( \sin A = \frac{1}{\sqrt{3}} \) and \( \sin B = \frac{1}{\sqrt{5}} \).
\( = \frac{\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}}}{\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}}} \)
To simplify, find a common denominator for the fractions in the numerator and denominator, which is \( \sqrt{3}\sqrt{5} = \sqrt{15} \).
Numerator: \( \frac{\sqrt{5}}{\sqrt{15}} + \frac{\sqrt{3}}{\sqrt{15}} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{15}} \)
Denominator: \( \frac{\sqrt{5}}{\sqrt{15}} - \frac{\sqrt{3}}{\sqrt{15}} = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{15}} \)
So, the expression becomes:
\( = \frac{\frac{\sqrt{5}+\sqrt{3}}{\sqrt{15}}}{\frac{\sqrt{5}-\sqrt{3}}{\sqrt{15}}} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, \( \sqrt{5}+\sqrt{3} \).
\( = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}} \)
\( = \frac{(\sqrt{5})^2 + (\sqrt{3})^2 + 2\sqrt{5}\sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2} \)
\( = \frac{5 + 3 + 2\sqrt{15}}{5 - 3} \)
\( = \frac{8 + 2\sqrt{15}}{2} \)
\( = \frac{2(4 + \sqrt{15})}{2} \)
\( = 4 + \sqrt{15} \)
Thus, the value of \( \tan \frac{1}{2}(A+B) \cot \frac{1}{2}(A-B) \) is \( 4 + \sqrt{15} \). This problem shows how to use half-angle product identities effectively.
In simple words: We want to find the value of an expression that has 'tan' and 'cot' of half-angles. We use a special formula that changes this expression into a ratio of the sum and difference of sin A and sin B. Then we plug in the given values for sin A and sin B and simplify the resulting fraction. We rationalize the denominator to get a clean final answer.

🎯 Exam Tip: Remember the identities that relate sums/differences of half-angles to single angles (e.g., \( \frac{\sin X + \sin Y}{\sin X - \sin Y} \)). This can greatly simplify calculations by allowing you to use the given values directly.

 

Question 14. If \( \sin \theta = n \sin (\theta + 2\alpha) \), show that \( (n - 1) \tan (\theta + \alpha) + (n + 1) \tan \alpha = 0 \).
Answer: We are given the relation \( \sin \theta = n \sin (\theta + 2\alpha) \). We need to show that \( (n-1) \tan (\theta + \alpha) + (n+1) \tan \alpha = 0 \).
From the given relation, we can write:
\( \frac{\sin \theta}{\sin (\theta + 2\alpha)} = n \)
To apply componendo and dividendo, we write \( n \) as \( \frac{n}{1} \):
\( \frac{\sin \theta}{\sin (\theta + 2\alpha)} = \frac{n}{1} \)
Applying componendo and dividendo, which states that if \( \frac{a}{b} = \frac{c}{d} \), then \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \):
\( \frac{\sin (\theta + 2\alpha) + \sin \theta}{\sin (\theta + 2\alpha) - \sin \theta} = \frac{n+1}{n-1} \)
Now, we use the sum-to-product formulas:
\( \sin C + \sin D = 2 \sin \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right) \)
\( \sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right) \)
For the numerator, let \( C = \theta + 2\alpha \) and \( D = \theta \):
\( \frac{C+D}{2} = \frac{\theta + 2\alpha + \theta}{2} = \frac{2\theta + 2\alpha}{2} = \theta + \alpha \)
\( \frac{C-D}{2} = \frac{\theta + 2\alpha - \theta}{2} = \frac{2\alpha}{2} = \alpha \)
So, the numerator becomes \( 2 \sin (\theta + \alpha) \cos \alpha \).
For the denominator, using the same \( C \) and \( D \):
\( \frac{C+D}{2} = \theta + \alpha \)
\( \frac{C-D}{2} = \alpha \)
So, the denominator becomes \( 2 \cos (\theta + \alpha) \sin \alpha \).
Substitute these back into the equation:
\( \frac{2 \sin (\theta + \alpha) \cos \alpha}{2 \cos (\theta + \alpha) \sin \alpha} = \frac{n+1}{n-1} \)
The 2's cancel out. We can rewrite the left side using tangent and cotangent:
\( \implies \left( \frac{\sin (\theta + \alpha)}{\cos (\theta + \alpha)} \right) \left( \frac{\cos \alpha}{\sin \alpha} \right) = \frac{n+1}{n-1} \)
\( \implies \tan (\theta + \alpha) \cot \alpha = \frac{n+1}{n-1} \)
Since \( \cot \alpha = \frac{1}{\tan \alpha} \), we have:
\( \implies \frac{\tan (\theta + \alpha)}{\tan \alpha} = \frac{n+1}{n-1} \)
Now, cross-multiply:
\( (n-1) \tan (\theta + \alpha) = (n+1) \tan \alpha \)
Rearrange the terms to match the required expression:
\( (n-1) \tan (\theta + \alpha) - (n+1) \tan \alpha = 0 \)
Wait, the target expression is \( (n - 1) \tan (\theta + \alpha) + (n + 1) \tan \alpha = 0 \). Let's recheck the step where we applied componendo and dividendo.
If \( \frac{a}{b} = \frac{c}{d} \), then \( \frac{b+a}{b-a} = \frac{d+c}{d-c} \) or \( \frac{a-b}{a+b} = \frac{c-d}{c+d} \).
Let's use \( \frac{\sin (\theta + 2\alpha)}{\sin \theta} = \frac{1}{n} \).
Then, applying componendo and dividendo as \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \):
\( \frac{\sin (\theta + 2\alpha) + \sin \theta}{\sin (\theta + 2\alpha) - \sin \theta} = \frac{1+n}{1-n} \) (This is what we had, just swapped sides)
\( \frac{2 \sin (\theta + \alpha) \cos \alpha}{2 \cos (\theta + \alpha) \sin \alpha} = \frac{n+1}{1-n} \)
\( \implies \tan (\theta + \alpha) \cot \alpha = \frac{n+1}{1-n} \)
\( \implies \frac{\tan (\theta + \alpha)}{\tan \alpha} = \frac{n+1}{1-n} \)
Cross-multiply:
\( (1-n) \tan (\theta + \alpha) = (n+1) \tan \alpha \)
\( \tan (\theta + \alpha) - n \tan (\theta + \alpha) = n \tan \alpha + \tan \alpha \)
Rearrange:
\( \tan (\theta + \alpha) - \tan \alpha = n \tan (\theta + \alpha) + n \tan \alpha \)
\( \tan (\theta + \alpha) - \tan \alpha = n (\tan (\theta + \alpha) + \tan \alpha) \)
This isn't leading to the desired result directly. Let's restart from \( \frac{\sin \theta}{\sin (\theta + 2\alpha)} = \frac{n}{1} \) and apply a slightly different form of componendo and dividendo or manipulate directly.
Given: \( \sin \theta = n \sin (\theta + 2\alpha) \)
\( \implies \frac{\sin (\theta + 2\alpha)}{\sin \theta} = \frac{1}{n} \)
Apply componendo and dividendo directly as \( \frac{\sin (\theta + 2\alpha) + \sin \theta}{\sin (\theta + 2\alpha) - \sin \theta} = \frac{1+n}{1-n} \).
This leads to: \( \frac{\tan (\theta + \alpha)}{\tan \alpha} = \frac{1+n}{1-n} \)
\( \implies (1-n) \tan (\theta + \alpha) = (1+n) \tan \alpha \)
\( \implies \tan (\theta + \alpha) - n \tan (\theta + \alpha) = \tan \alpha + n \tan \alpha \)
\( \implies \tan (\theta + \alpha) - \tan \alpha = n \tan (\theta + \alpha) + n \tan \alpha \)
\( \implies \tan (\theta + \alpha) - n \tan (\theta + \alpha) - \tan \alpha - n \tan \alpha = 0 \)
\( \implies (1-n) \tan (\theta + \alpha) - (1+n) \tan \alpha = 0 \)
This is \( (1-n) \tan (\theta + \alpha) - (n+1) \tan \alpha = 0 \).
Multiplying the entire equation by -1 will give the desired form:
\( - (1-n) \tan (\theta + \alpha) + (n+1) \tan \alpha = 0 \)
\( (n-1) \tan (\theta + \alpha) + (n+1) \tan \alpha = 0 \)
Thus, the identity is proven. The key was to correctly apply componendo and dividendo and then manipulate the signs carefully. This illustrates how algebraic rearrangement is often as important as trigonometric identities.
In simple words: We start with the given equation and rearrange it to form a ratio. Then, we apply a mathematical rule called componendo and dividendo. This transforms the sines into an expression involving tangents of sums and differences. After simplifying and cross-multiplying, we rearrange the terms to match the equation we needed to prove, making sure to handle the positive and negative signs correctly.

🎯 Exam Tip: When given an equation of the form \( \sin A = n \sin B \), a common strategy is to rearrange it as a ratio \( \frac{\sin A}{\sin B} = n \) and then apply componendo and dividendo. Be meticulous with algebraic manipulation and signs.

 

Question 15. If \( \tan \frac { \alpha }{ 2 } \) and \( \tan \frac { \beta }{ 2 } \) are the roots of the equation \( 8x^2 – 26x + 15 = 0 \), then find the value of \( \cos (\alpha + \beta) \).
Answer: We are given that \( \tan \frac{\alpha}{2} \) and \( \tan \frac{\beta}{2} \) are the roots of the quadratic equation \( 8x^2 - 26x + 15 = 0 \). We need to find the value of \( \cos (\alpha + \beta) \).
Let \( t_1 = \tan \frac{\alpha}{2} \) and \( t_2 = \tan \frac{\beta}{2} \).
For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is \( - \frac{b}{a} \) and the product of the roots is \( \frac{c}{a} \).
From \( 8x^2 - 26x + 15 = 0 \):
Sum of roots: \( t_1 + t_2 = \tan \frac{\alpha}{2} + \tan \frac{\beta}{2} = - \frac{-26}{8} = \frac{26}{8} = \frac{13}{4} \)
Product of roots: \( t_1 t_2 = \tan \frac{\alpha}{2} \tan \frac{\beta}{2} = \frac{15}{8} \)
Now we need to find \( \cos (\alpha + \beta) \). We use the tangent half-angle formula for \( \cos (A+B) \):
\( \cos (A+B) = \frac{1 - \tan^2 \left( \frac{A+B}{2} \right)}{1 + \tan^2 \left( \frac{A+B}{2} \right)} \)
So, \( \cos (\alpha + \beta) = \frac{1 - \tan^2 \left( \frac{\alpha+\beta}{2} \right)}{1 + \tan^2 \left( \frac{\alpha+\beta}{2} \right)} \)
First, let's find \( \tan \left( \frac{\alpha+\beta}{2} \right) \). We use the tangent addition formula:
\( \tan \left( \frac{\alpha+\beta}{2} \right) = \frac{\tan \frac{\alpha}{2} + \tan \frac{\beta}{2}}{1 - \tan \frac{\alpha}{2} \tan \frac{\beta}{2}} \)
Substitute the sum and product of the roots we found:
\( \tan \left( \frac{\alpha+\beta}{2} \right) = \frac{\frac{13}{4}}{1 - \frac{15}{8}} \)
Calculate the denominator:
\( 1 - \frac{15}{8} = \frac{8}{8} - \frac{15}{8} = \frac{8-15}{8} = \frac{-7}{8} \)
So, \( \tan \left( \frac{\alpha+\beta}{2} \right) = \frac{\frac{13}{4}}{\frac{-7}{8}} \)
\( = \frac{13}{4} \times \frac{8}{-7} = \frac{13 \times 2}{-7} = \frac{-26}{7} \)
Now, we have \( \tan \left( \frac{\alpha+\beta}{2} \right) = \frac{-26}{7} \).
Next, substitute this into the formula for \( \cos (\alpha + \beta) \):
\( \cos (\alpha + \beta) = \frac{1 - \left( \frac{-26}{7} \right)^2}{1 + \left( \frac{-26}{7} \right)^2} \)
\( \left( \frac{-26}{7} \right)^2 = \frac{(-26)^2}{7^2} = \frac{676}{49} \)
\( \cos (\alpha + \beta) = \frac{1 - \frac{676}{49}}{1 + \frac{676}{49}} \)
Calculate the numerator: \( 1 - \frac{676}{49} = \frac{49}{49} - \frac{676}{49} = \frac{49 - 676}{49} = \frac{-627}{49} \)
Calculate the denominator: \( 1 + \frac{676}{49} = \frac{49}{49} + \frac{676}{49} = \frac{49 + 676}{49} = \frac{725}{49} \)
So, \( \cos (\alpha + \beta) = \frac{\frac{-627}{49}}{\frac{725}{49}} \)
The \( 49 \) in the denominators cancel out:
\( \cos (\alpha + \beta) = \frac{-627}{725} \)
This is the value of \( \cos (\alpha + \beta) \). This problem effectively combines knowledge of quadratic equations with trigonometric identities.
In simple words: First, we find the sum and product of the roots of the given quadratic equation. These roots are tan of alpha/2 and tan of beta/2. Then, we use the sum and product to find tan of (alpha + beta)/2. Finally, we use a special formula that connects tan of a half-angle to the cosine of the full angle to get our answer for cos (alpha + beta).

🎯 Exam Tip: This type of problem often links quadratic equations with trigonometric identities. Remember the sum and product of roots formulas for quadratics, and use the half-angle formulas (especially for tangent in terms of cosine) to bridge the gap.

 

Question 16. Prove that \( \left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n= \begin{cases}2 \cot ^n\left(\frac{A-B}{2}\right) &, \text { if } n \text { is even. } \\ 0 & \text {,if } n \text { is odd. }\end{cases} \)
Answer: Let's simplify each term in the left-hand side (L.H.S.) separately.
First term: \( \frac{\cos A+\cos B}{\sin A-\sin B} \)
Using sum-to-product formulas:
\( \cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \)
\( \sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \)
So, \( \frac{\cos A+\cos B}{\sin A-\sin B} = \frac{2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)}{2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)} \)
Canceling \( 2 \cos \left( \frac{A+B}{2} \right) \), we get:
\( \frac{\cos \left( \frac{A-B}{2} \right)}{\sin \left( \frac{A-B}{2} \right)} = \cot \left( \frac{A-B}{2} \right) \)
Second term: \( \frac{\sin A+\sin B}{\cos A-\cos B} \)
Using sum-to-product formulas:
\( \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \)
\( \cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \)
So, \( \frac{\sin A+\sin B}{\cos A-\cos B} = \frac{2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)}{-2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)} \)
Canceling \( 2 \sin \left( \frac{A+B}{2} \right) \), we get:
\( \frac{\cos \left( \frac{A-B}{2} \right)}{-\sin \left( \frac{A-B}{2} \right)} = - \cot \left( \frac{A-B}{2} \right) \)
Now, substitute these simplified terms back into the L.H.S. expression:
L.H.S. \( = \left[ \cot \left( \frac{A-B}{2} \right) \right]^n + \left[ - \cot \left( \frac{A-B}{2} \right) \right]^n \)
L.H.S. \( = \cot^n \left( \frac{A-B}{2} \right) + (-1)^n \cot^n \left( \frac{A-B}{2} \right) \)
Now, we consider two cases for \( n \):
Case I: If \( n \) is an even number.
If \( n \) is even, then \( (-1)^n = 1 \).
L.H.S. \( = \cot^n \left( \frac{A-B}{2} \right) + 1 \cdot \cot^n \left( \frac{A-B}{2} \right) \)
L.H.S. \( = 2 \cot^n \left( \frac{A-B}{2} \right) \)
Case II: If \( n \) is an odd number.
If \( n \) is odd, then \( (-1)^n = -1 \).
L.H.S. \( = \cot^n \left( \frac{A-B}{2} \right) + (-1) \cot^n \left( \frac{A-B}{2} \right) \)
L.H.S. \( = \cot^n \left( \frac{A-B}{2} \right) - \cot^n \left( \frac{A-B}{2} \right) \)
L.H.S. \( = 0 \)
Thus, the given identity is proven for both even and odd values of \( n \). This problem is a comprehensive test of sum-to-product identities and properties of exponents.
In simple words: We simplify each fraction on the left side of the equation separately using formulas that change sums of sines or cosines into products. After simplifying, we get cotangent of (A-B)/2 for the first term and negative cotangent of (A-B)/2 for the second term. Then we look at what happens when these terms are raised to the power 'n'. If 'n' is an even number, the negative sign disappears, and the two terms add up. If 'n' is an odd number, the negative sign stays, and the two terms cancel each other out, giving zero.

🎯 Exam Tip: This problem requires a careful application of sum-to-product formulas. Pay close attention to the signs, especially when applying the \( \cos A - \cos B \) formula, which has a negative sign. Also, remember how negative bases behave when raised to even or odd powers.

 

Question 17. Find \( \sin \frac { x }{ 2 } \), \( \cos \frac { x }{ 2 } \) and \( \tan \frac { x }{ 2 } \) in each of the following cases : (i) \( \sin x = \frac { 1 }{ 4 } \), \( x \) in II quadrant, (ii) \( \tan x = \frac { -4 }{ 3 } \), \( x \) in II quadrant.
Answer:
(i) Given \( \sin x = \frac{1}{4} \) and \( x \) is in the second quadrant.
If \( x \) is in the second quadrant, it means \( 90^\circ < x < 180^\circ \).
Dividing by 2, we get \( \frac{90^\circ}{2} < \frac{x}{2} < \frac{180^\circ}{2} \), which means \( 45^\circ < \frac{x}{2} < 90^\circ \).
This places \( \frac{x}{2} \) in the first quadrant. In the first quadrant, \( \sin \frac{x}{2} \), \( \cos \frac{x}{2} \), and \( \tan \frac{x}{2} \) are all positive.
First, find \( \cos x \). Since \( x \) is in the second quadrant, \( \cos x \) is negative.
\( \cos^2 x = 1 - \sin^2 x = 1 - \left( \frac{1}{4} \right)^2 = 1 - \frac{1}{16} = \frac{15}{16} \)
\( \cos x = - \sqrt{\frac{15}{16}} = - \frac{\sqrt{15}}{4} \)
Now, use the half-angle formulas:
\( \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} \)
\( \sin^2 \frac{x}{2} = \frac{1 - \left( - \frac{\sqrt{15}}{4} \right)}{2} = \frac{1 + \frac{\sqrt{15}}{4}}{2} = \frac{\frac{4+\sqrt{15}}{4}}{2} = \frac{4+\sqrt{15}}{8} \)
Since \( \frac{x}{2} \) is in the first quadrant, \( \sin \frac{x}{2} \) is positive:
\( \sin \frac{x}{2} = \sqrt{\frac{4+\sqrt{15}}{8}} = \frac{\sqrt{4+\sqrt{15}}}{\sqrt{8}} = \frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}} \)
To rationalize, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \): \( \sin \frac{x}{2} = \frac{\sqrt{2(4+\sqrt{15})}}{4} = \frac{\sqrt{8+2\sqrt{15}}}{4} \)
Next, for \( \cos \frac{x}{2} \):
\( \cos^2 \frac{x}{2} = \frac{1 + \cos x}{2} \)
\( \cos^2 \frac{x}{2} = \frac{1 + \left( - \frac{\sqrt{15}}{4} \right)}{2} = \frac{1 - \frac{\sqrt{15}}{4}}{2} = \frac{\frac{4-\sqrt{15}}{4}}{2} = \frac{4-\sqrt{15}}{8} \)
Since \( \frac{x}{2} \) is in the first quadrant, \( \cos \frac{x}{2} \) is positive:
\( \cos \frac{x}{2} = \sqrt{\frac{4-\sqrt{15}}{8}} = \frac{\sqrt{4-\sqrt{15}}}{\sqrt{8}} = \frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}} \)
To rationalize: \( \cos \frac{x}{2} = \frac{\sqrt{2(4-\sqrt{15})}}{4} = \frac{\sqrt{8-2\sqrt{15}}}{4} \)
Finally, for \( \tan \frac{x}{2} \):
\( \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\sqrt{\frac{4+\sqrt{15}}{8}}}{\sqrt{\frac{4-\sqrt{15}}{8}}} = \sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}} \)
To rationalize, multiply by \( \frac{4+\sqrt{15}}{4+\sqrt{15}} \) inside the square root:
\( \tan \frac{x}{2} = \sqrt{\frac{(4+\sqrt{15})^2}{4^2 - (\sqrt{15})^2}} = \sqrt{\frac{(4+\sqrt{15})^2}{16-15}} = \sqrt{\frac{(4+\sqrt{15})^2}{1}} = 4+\sqrt{15} \)
Alternatively, using the formula \( \tan \frac{x}{2} = \frac{1-\cos x}{\sin x} \):
\( \tan \frac{x}{2} = \frac{1 - (-\frac{\sqrt{15}}{4})}{\frac{1}{4}} = \frac{1 + \frac{\sqrt{15}}{4}}{\frac{1}{4}} = \frac{\frac{4+\sqrt{15}}{4}}{\frac{1}{4}} = 4+\sqrt{15} \)
This result is consistent with the square root calculation. These calculations show how to use half-angle formulas and handle signs based on the quadrant.
(ii) Given \( \tan x = - \frac{4}{3} \) and \( x \) is in the second quadrant.
Again, if \( x \) is in the second quadrant \( (90^\circ < x < 180^\circ) \), then \( \frac{x}{2} \) is in the first quadrant \( (45^\circ < \frac{x}{2} < 90^\circ) \).
So, \( \sin \frac{x}{2} \), \( \cos \frac{x}{2} \), and \( \tan \frac{x}{2} \) are all positive.
First, find \( \sec x \). \( \sec^2 x = 1 + \tan^2 x = 1 + \left( - \frac{4}{3} \right)^2 = 1 + \frac{16}{9} = \frac{9+16}{9} = \frac{25}{9} \)
\( \sec x = \pm \sqrt{\frac{25}{9}} = \pm \frac{5}{3} \).
Since \( x \) is in the second quadrant, \( \cos x \) is negative, so \( \sec x \) is also negative.
\( \sec x = - \frac{5}{3} \).
From this, \( \cos x = \frac{1}{\sec x} = - \frac{3}{5} \).
Now, use the half-angle formulas:
\( \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} = \frac{1 - \left( - \frac{3}{5} \right)}{2} = \frac{1 + \frac{3}{5}}{2} = \frac{\frac{5+3}{5}}{2} = \frac{\frac{8}{5}}{2} = \frac{8}{10} = \frac{4}{5} \)
Since \( \frac{x}{2} \) is in the first quadrant, \( \sin \frac{x}{2} \) is positive:
\( \sin \frac{x}{2} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \)
Next, for \( \cos \frac{x}{2} \):
\( \cos^2 \frac{x}{2} = \frac{1 + \cos x}{2} = \frac{1 + \left( - \frac{3}{5} \right)}{2} = \frac{1 - \frac{3}{5}}{2} = \frac{\frac{5-3}{5}}{2} = \frac{\frac{2}{5}}{2} = \frac{2}{10} = \frac{1}{5} \)
Since \( \frac{x}{2} \) is in the first quadrant, \( \cos \frac{x}{2} \) is positive:
\( \cos \frac{x}{2} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \)
Finally, for \( \tan \frac{x}{2} \):
\( \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2 \)
Alternatively, using the formula \( \tan \frac{x}{2} = \frac{1-\cos x}{\sin x} \). First find \( \sin x \).
Since \( x \) is in QII and \( \tan x = -\frac{4}{3} \), we can imagine a right triangle with opposite side 4 and adjacent side 3. The hypotenuse would be 5. In QII, \( \sin x \) is positive.
So, \( \sin x = \frac{4}{5} \).
\( \tan \frac{x}{2} = \frac{1 - (-\frac{3}{5})}{\frac{4}{5}} = \frac{1 + \frac{3}{5}}{\frac{4}{5}} = \frac{\frac{8}{5}}{\frac{4}{5}} = \frac{8}{4} = 2 \)
This confirms the result. These steps show how to derive half-angle values from a given tangent and quadrant.
In simple words: First, we determine which quadrant x/2 falls into, which tells us if sine, cosine, and tangent of x/2 should be positive or negative. Then, we use the given sine or tan value for x to find cos x. Finally, we apply the half-angle formulas for sine, cosine, and tangent using the value of cos x to calculate sin x/2, cos x/2, and tan x/2.

🎯 Exam Tip: Always determine the quadrant of \( \frac{x}{2} \) first, as this dictates the sign of \( \sin \frac{x}{2} \), \( \cos \frac{x}{2} \), and \( \tan \frac{x}{2} \). Remember that \( \cos x \) will be negative in Q2. Also, know the alternative formulas for \( \tan \frac{x}{2} \), like \( \frac{1-\cos x}{\sin x} \) or \( \frac{\sin x}{1+\cos x} \), as they can sometimes simplify calculations.

 

Question 18. Prove that \( \cos 6x = 32 \cos^6 x – 48 \cos^4 x + 18 \cos^2 x − 1 \).
Answer: We need to prove the given identity starting from the left-hand side (L.H.S.).
L.H.S. \( = \cos 6x \)
We can write \( \cos 6x \) as \( \cos (2 \times 3x) \).
Using the double angle formula \( \cos 2\theta = 2 \cos^2 \theta - 1 \), let \( \theta = 3x \).
\( \implies \cos 6x = 2 \cos^2 (3x) - 1 \)
Now, we need to express \( \cos 3x \) in terms of \( \cos x \). Use the triple angle formula:
\( \cos 3x = 4 \cos^3 x - 3 \cos x \)
Substitute this into the expression for \( \cos 6x \):
\( \cos 6x = 2 (4 \cos^3 x - 3 \cos x)^2 - 1 \)
Expand the squared term \( (4 \cos^3 x - 3 \cos x)^2 \). This is in the form \( (a-b)^2 = a^2 - 2ab + b^2 \).
Let \( a = 4 \cos^3 x \) and \( b = 3 \cos x \).
\( (4 \cos^3 x - 3 \cos x)^2 = (4 \cos^3 x)^2 - 2(4 \cos^3 x)(3 \cos x) + (3 \cos x)^2 \)
\( = 16 \cos^6 x - 24 \cos^4 x + 9 \cos^2 x \)
Now, substitute this back into the expression for \( \cos 6x \):
\( \cos 6x = 2 (16 \cos^6 x - 24 \cos^4 x + 9 \cos^2 x) - 1 \)
Distribute the 2:
\( \cos 6x = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1 \)
This is equal to the right-hand side (R.H.S.).
Thus, the identity is proven. This demonstrates how to build up higher multiple angle formulas from simpler ones.
In simple words: We start with cos 6x and rewrite it as cos (2 times 3x). Then we use the double angle formula for cosine. Next, we replace cos 3x with its formula in terms of cos x. After carefully expanding the squared term and distributing the 2, we get the required expression that only has powers of cos x.

🎯 Exam Tip: When proving identities involving higher multiple angles (like \( \cos 6x \)), break them down using double or triple angle formulas. For example, \( \cos 6x = \cos (2 \cdot 3x) \) or \( \cos 6x = \cos (3 \cdot 2x) \). Choose the path that seems algebraically simpler.

 

Question 19. Prove that \( \sin^2 72^{\circ} – \sin^2 60^{\circ} = \frac{\sqrt{5}-1}{8} \).
Answer: We need to prove the given identity starting from the left-hand side (L.H.S.).
L.H.S. \( = \sin^2 72^{\circ} - \sin^2 60^{\circ} \)
We use the identity \( \sin^2 A - \sin^2 B = \sin (A+B) \sin (A-B) \).
Here, \( A = 72^{\circ} \) and \( B = 60^{\circ} \).
\( A+B = 72^{\circ} + 60^{\circ} = 132^{\circ} \)
\( A-B = 72^{\circ} - 60^{\circ} = 12^{\circ} \)
So, L.H.S. \( = \sin 132^{\circ} \sin 12^{\circ} \).
We know that \( \sin 132^{\circ} = \sin (180^{\circ} - 48^{\circ}) = \sin 48^{\circ} \).
So, L.H.S. \( = \sin 48^{\circ} \sin 12^{\circ} \).
Now, use the product-to-sum formula \( 2 \sin A \sin B = \cos (A-B) - \cos (A+B) \).
Multiply and divide by 2:
L.H.S. \( = \frac{1}{2} (2 \sin 48^{\circ} \sin 12^{\circ}) \)
\( = \frac{1}{2} [\cos (48^{\circ} - 12^{\circ}) - \cos (48^{\circ} + 12^{\circ})] \)
\( = \frac{1}{2} [\cos 36^{\circ} - \cos 60^{\circ}] \)
We know the exact values for \( \cos 36^{\circ} \) and \( \cos 60^{\circ} \):
\( \cos 36^{\circ} = \frac{\sqrt{5}+1}{4} \)
\( \cos 60^{\circ} = \frac{1}{2} \)
Substitute these values:
L.H.S. \( = \frac{1}{2} \left[ \frac{\sqrt{5}+1}{4} - \frac{1}{2} \right] \)
Find a common denominator inside the bracket:
\( = \frac{1}{2} \left[ \frac{\sqrt{5}+1}{4} - \frac{2}{4} \right] \)
\( = \frac{1}{2} \left[ \frac{\sqrt{5}+1-2}{4} \right] \)
\( = \frac{1}{2} \left[ \frac{\sqrt{5}-1}{4} \right] \)
\( = \frac{\sqrt{5}-1}{8} \)
This is equal to the right-hand side (R.H.S.).
Thus, the identity is proven. This shows the importance of knowing special trigonometric values for common angles and using product-to-sum identities.
In simple words: We start with the left side of the equation, which has a difference of two sine squared terms. We use a formula that changes this into a product of sines. We simplify the angles and then use another formula that changes a product of sines into a difference of cosines. Finally, we plug in the exact known values for cos 36 degrees and cos 60 degrees, and simplify the expression to get the right side of the equation.

🎯 Exam Tip: Familiarize yourself with exact trigonometric values for special angles like \( 36^\circ, 54^\circ, 72^\circ \), etc., in addition to the standard \( 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ \). The identity \( \sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B) \) is also very useful for these types of problems.

 

Question 20. If \( \tan x, \tan y, \tan z \) are in GP., show that \( \cos 2y = \frac{\cos (x+z)}{\cos (x-z)} \).
Answer: We are given that \( \tan x, \tan y, \tan z \) are in Geometric Progression (GP.). This means that the ratio of consecutive terms is constant.
So, \( \frac{\tan y}{\tan x} = \frac{\tan z}{\tan y} \)
\( \implies \tan^2 y = \tan x \tan z \)
We can express tangent in terms of sine and cosine: \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
So, \( \frac{\sin^2 y}{\cos^2 y} = \frac{\sin x}{\cos x} \cdot \frac{\sin z}{\cos z} \)
\( \implies \frac{\sin^2 y}{\cos^2 y} = \frac{\sin x \sin z}{\cos x \cos z} \)
Now, we want to prove \( \cos 2y = \frac{\cos (x+z)}{\cos (x-z)} \). Let's work with the R.H.S.
R.H.S. \( = \frac{\cos (x+z)}{\cos (x-z)} \)
Using the cosine sum and difference formulas:
\( \cos (x+z) = \cos x \cos z - \sin x \sin z \)
\( \cos (x-z) = \cos x \cos z + \sin x \sin z \)
So, R.H.S. \( = \frac{\cos x \cos z - \sin x \sin z}{\cos x \cos z + \sin x \sin z} \)
Now, divide both the numerator and the denominator by \( \cos x \cos z \):
\( = \frac{\frac{\cos x \cos z}{\cos x \cos z} - \frac{\sin x \sin z}{\cos x \cos z}}{\frac{\cos x \cos z}{\cos x \cos z} + \frac{\sin x \sin z}{\cos x \cos z}} \)
\( = \frac{1 - \tan x \tan z}{1 + \tan x \tan z} \)
From the GP condition, we know \( \tan x \tan z = \tan^2 y \). Substitute this into the R.H.S.:
R.H.S. \( = \frac{1 - \tan^2 y}{1 + \tan^2 y} \)
This expression is the formula for \( \cos 2y \) in terms of \( \tan y \).
\( \cos 2y = \frac{1 - \tan^2 y}{1 + \tan^2 y} \)
So, R.H.S. \( = \cos 2y \).
Since L.H.S. \( = \cos 2y \) and R.H.S. \( = \cos 2y \), the identity is proven. This shows a nice connection between geometric progressions and trigonometric identities.
In simple words: Since tan x, tan y, and tan z are in a geometric progression, the square of tan y is equal to the product of tan x and tan z. We start with the right side of the equation we need to prove. We expand the cosine sum and difference terms and then divide everything by cos x cos z to get an expression with tan x tan z. We then replace tan x tan z with tan squared y, which directly gives us the formula for cos 2y, matching the left side.

🎯 Exam Tip: When terms are in GP, always start by writing the defining property: the square of the middle term equals the product of the first and last terms. Then, try to manipulate the more complex side of the identity to be proven, looking for opportunities to substitute this GP relation.

 

Question 21. If \( \tan \frac { \alpha }{ 2 } : \tan \frac { \beta }{ 2 } = 1 : \sqrt{3} \) show that \( \cos B = \frac{2 \cos \alpha-1}{2-\cos \alpha} \).
Answer: We are given \( \tan \frac{\alpha}{2} : \tan \frac{\beta}{2} = 1 : \sqrt{3} \). This can be written as:
\( \frac{\tan \frac{\alpha}{2}}{\tan \frac{\beta}{2}} = \frac{1}{\sqrt{3}} \)
Squaring both sides:
\( \left( \frac{\tan \frac{\alpha}{2}}{\tan \frac{\beta}{2}} \right)^2 = \left( \frac{1}{\sqrt{3}} \right)^2 \)
\( \implies \frac{\tan^2 \frac{\alpha}{2}}{\tan^2 \frac{\beta}{2}} = \frac{1}{3} \)
Now, we want to show \( \cos \beta = \frac{2 \cos \alpha - 1}{2 - \cos \alpha} \).
We use the half-angle formula for cosine: \( \cos \theta = \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} \).
From this, we can derive expressions for \( \tan^2 \frac{\theta}{2} \):
\( \cos \theta (1 + \tan^2 \frac{\theta}{2}) = 1 - \tan^2 \frac{\theta}{2} \)
\( \cos \theta + \cos \theta \tan^2 \frac{\theta}{2} = 1 - \tan^2 \frac{\theta}{2} \)
\( \tan^2 \frac{\theta}{2} ( \cos \theta + 1) = 1 - \cos \theta \)
\( \implies \tan^2 \frac{\theta}{2} = \frac{1 - \cos \theta}{1 + \cos \theta} \)
Now, substitute this into our ratio:
\( \frac{\frac{1 - \cos \alpha}{1 + \cos \alpha}}{\frac{1 - \cos \beta}{1 + \cos \beta}} = \frac{1}{3} \)
\( \implies \frac{1 - \cos \alpha}{1 + \cos \alpha} \cdot \frac{1 + \cos \beta}{1 - \cos \beta} = \frac{1}{3} \)
Cross-multiply:
\( 3 (1 - \cos \alpha) (1 + \cos \beta) = 1 (1 + \cos \alpha) (1 - \cos \beta) \)
\( 3 (1 + \cos \beta - \cos \alpha - \cos \alpha \cos \beta) = 1 - \cos \beta + \cos \alpha - \cos \alpha \cos \beta \)
\( 3 + 3 \cos \beta - 3 \cos \alpha - 3 \cos \alpha \cos \beta = 1 - \cos \beta + \cos \alpha - \cos \alpha \cos \beta \)
Our goal is to isolate \( \cos \beta \). Let's gather all terms with \( \cos \beta \) on one side and the rest on the other side.
\( 3 \cos \beta + \cos \beta - 3 \cos \alpha \cos \beta + \cos \alpha \cos \beta = 1 + 3 \cos \alpha - 3 \)
\( 4 \cos \beta - 2 \cos \alpha \cos \beta = 3 \cos \alpha - 2 \)
Factor out \( \cos \beta \) from the left side:
\( \cos \beta (4 - 2 \cos \alpha) = 3 \cos \alpha - 2 \)
Now, divide to solve for \( \cos \beta \):
\( \cos \beta = \frac{3 \cos \alpha - 2}{4 - 2 \cos \alpha} \)
We can factor out a 1 from the numerator to make it look like \( 2 \cos \alpha - 1 \) if we swap signs, and factor a 2 from the denominator.
\( \cos \beta = \frac{-(2 - 3 \cos \alpha)}{2(2 - \cos \alpha)} \) - this is not directly matching. Let's recheck rearrangement.
From \( 4 \cos \beta - 2 \cos \alpha \cos \beta = 3 \cos \alpha - 2 \):
Let's rearrange the terms in the numerator to match \( 2 \cos \alpha - 1 \).
The numerator is \( 3 \cos \alpha - 2 \). This can be written as \( 2 \cos \alpha - 1 + \cos \alpha - 1 \). No.
Let's divide numerator and denominator by -1 to get positive terms where possible.
\( \cos \beta = \frac{2 - 3 \cos \alpha}{2 \cos \alpha - 4} \)
This is still not matching \( \frac{2 \cos \alpha - 1}{2 - \cos \alpha} \). Let's restart from \( 3 + 3 \cos \beta - 3 \cos \alpha - 3 \cos \alpha \cos \beta = 1 - \cos \beta + \cos \alpha - \cos \alpha \cos \beta \) and collect terms differently.
Move all \( \cos \beta \) terms to one side, and all other terms to the other.
\( 3 \cos \beta + \cos \beta = 1 - 3 + \cos \alpha + 3 \cos \alpha - \cos \alpha \cos \beta + 3 \cos \alpha \cos \beta \)
\( 4 \cos \beta = -2 + 4 \cos \alpha + 2 \cos \alpha \cos \beta \)
Move the \( 2 \cos \alpha \cos \beta \) term to the left:
\( 4 \cos \beta - 2 \cos \alpha \cos \beta = 4 \cos \alpha - 2 \)
Factor out \( \cos \beta \) from the left side:
\( \cos \beta (4 - 2 \cos \alpha) = 4 \cos \alpha - 2 \)
Now, divide by \( (4 - 2 \cos \alpha) \):
\( \cos \beta = \frac{4 \cos \alpha - 2}{4 - 2 \cos \alpha} \)
Factor out 2 from both the numerator and the denominator:
\( \cos \beta = \frac{2(2 \cos \alpha - 1)}{2(2 - \cos \alpha)} \)
Cancel the 2's:
\( \cos \beta = \frac{2 \cos \alpha - 1}{2 - \cos \alpha} \)
This matches the required expression. Thus, the identity is proven. This problem emphasizes careful algebraic manipulation after applying trigonometric identities.
In simple words: We start by writing the given ratio of tangents. Then, we square both sides and replace the squared tangent terms with their equivalent formulas that involve cosine. We rearrange the resulting equation to solve for cos beta. After careful algebraic steps like cross-multiplying, collecting similar terms, and factoring, we simplify the expression to get the required formula for cos beta.

🎯 Exam Tip: When dealing with ratios of half-angle tangents, using the identity \( \tan^2 \frac{\theta}{2} = \frac{1-\cos \theta}{1+\cos \theta} \) is often the most direct approach. Be very careful and systematic with the algebraic rearrangement to avoid sign errors or missing common factors.

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