Question

If \(x = \sin(2\tan^{-1}2)\), \(y = \sin\left(\frac{1}{2}\tan^{-1}\frac{4}{3}\right)\), then:

• A. \(x = 1 - y\)
• B. \(x^2 = 1 - y\)
• C. \(x^2 = 1 + y\)
• D. \(y^2 = 1 - x\)

Hint:

Convert \(\tan^{-1}\) values into right triangles to quickly get \(\sin\theta, \cos\theta\).

Answer 1

The Correct Option is D

Concept: Use standard identities

Step 1:Evaluate \(x\) Let \(\theta = \tan^{-1}2\), so \(\tan\theta = 2\) \[ \sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2 \cdot 2}{1 + 4} = \frac{4}{5} \] \[ x = \frac{4}{5} \]

Step 2:Evaluate \(y\) Let \(\phi = \tan^{-1}\frac{4}{3}\), so \(\tan\phi = \frac{4}{3}\) Using triangle: \[ \sin\phi = \frac{4}{5}, \quad \cos\phi = \frac{3}{5} \] Half-angle formula: \[ y = \sin\frac{\phi}{2} = \sqrt{\frac{1 - \cos\phi}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{2/5}{2}} = \frac{1}{\sqrt{5}} \]

Step 3:Verify relation \[ y^2 = \frac{1}{5}, \quad 1 - x = 1 - \frac{4}{5} = \frac{1}{5} \] \[ \Rightarrow y^2 = 1 - x \]