Question

If \( \frac{1}{2} \sin^{-1} \left( \frac{3\sin 2\theta}{5+4\cos 2\theta} \right) = \tan^{-1} x \) then \( x = \)

• A. \( \tan \frac{\theta}{3} \)
• B. \( \frac{1}{3} \tan \theta \)
• C. \( \tan 3\theta \)
• D. \( \frac{1}{3} \tan 3\theta \) Correct Answer

Hint:

When dealing with inverse trigonometric functions, convert them to direct trigonometric functions. Use substitutions like \( \tan\theta = t \). The identity \( \sin(2A) = \frac{2\tan A}{1+\tan^2 A} \) is very useful. If an expression like \( \frac{k \cdot f(y)}{f(y)^2 + k^2} \) appears, it might be related to \( \sin(2 \tan^{-1}(f(y)/k)) \) or similar forms.

Answer 1

The Correct Option is B

Step 1: Simplify the argument of \( \sin^{-1} \).
Let \( A = \frac{3\sin 2\theta}{5+4\cos 2\theta} \).
Use \( \sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} \) and \( \cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} \).
Let \( t = \tan\theta \).
\[ A = \frac{3\left(\frac{2t}{1+t^2}\right)}{5+4\left(\frac{1-t^2}{1+t^2}\right)} = \frac{\frac{6t}{1+t^2}}{\frac{5(1+t^2)+4(1-t^2)}{1+t^2}} = \frac{6t}{5+5t^2+4-4t^2} = \frac{6t}{t^2+9} \]
Step 2: Let \( \frac{1}{2} \sin^{-1} A = \alpha \).
Then \( \sin^{-1} A = 2\alpha \), so \( \sin(2\alpha) = A \).
We are given \( \alpha = \tan^{-1} x \), so \( \tan \alpha = x \).
We know \( \sin(2\alpha) = \frac{2\tan\alpha}{1+\tan^2\alpha} = \frac{2x}{1+x^2} \).
So, \( \frac{2x}{1+x^2} = A = \frac{6t}{t^2+9} \).
\[ \frac{2x}{1+x^2} = \frac{6\tan\theta}{\tan^2\theta+9} \]
Step 3: Try to match the form.
We are looking for \( x \) in terms of \( \tan\theta \).
If we assume \( x = k \tan\theta = kt \), then \[ \frac{2kt}{1+(kt)^2} = \frac{6t}{t^2+9} \] If \( t \ne 0 \), we can cancel \( t \).
\[ \frac{2k}{1+k^2t^2} = \frac{6}{t^2+9} \] This must hold for all \( \theta \) (or \( t \)).
Comparing numerators, \( 2k=6 \implies k=3 \).
(This suggests \(x = 3 \tan \theta\)) Substitute \(k=3\): \[ \frac{6}{1+9t^2} = \frac{6}{t^2+9} \] This means \( 1+9t^2 = t^2+9 \implies 8t^2 = 8 \implies t^2=1 \implies \tan^2\theta=1 \).
This is not general.
Alternative approach for Step 2: Let \( \tan^{-1} x = \phi \).
Then \( x = \tan\phi \).
The equation is \( \frac{1}{2} \sin^{-1} \left( \frac{6\tan\theta}{\tan^2\theta+9} \right) = \phi \).
So \( \sin^{-1} \left( \frac{6\tan\theta}{\tan^2\theta+9} \right) = 2\phi \).
Taking sine on both sides: \( \frac{6\tan\theta}{\tan^2\theta+9} = \sin(2\phi) = \frac{2\tan\phi}{1+\tan^2\phi} \).
\[ \frac{3\tan\theta}{\tan^2\theta+9} = \frac{\tan\phi}{1+\tan^2\phi} \] Let \( \tan\theta = t_1 \) and \( \tan\phi = t_2 (=x) \).
\[ \frac{3t_1}{t_1^2+9} = \frac{t_2}{1+t_2^2} \] We need to find \( t_2 \) in terms of \( t_1 \).
Consider the substitution \( t_1 = 3T \).
(This comes from seeing the 9 with \(t_1^2\)) \[ \frac{3(3T)}{(3T)^2+9} = \frac{9T}{9T^2+9} = \frac{9T}{9(T^2+1)} = \frac{T}{T^2+1} \] So we have \( \frac{T}{T^2+1} = \frac{t_2}{1+t_2^2} \).
This implies \( t_2 = T \).
We set \( t_1 = 3T \), so \( \tan\theta = 3T \).
Then \( T = \frac{1}{3}\tan\theta \).
Since \( t_2 = T \), we have \( x = \tan\phi = T = \frac{1}{3}\tan\theta \).
This matches option (2).