\cot ^ - 1( \sqrt \cos \alpha ) - \tan ^ - 1( \sqrt \cos \alpha ) = x, then sin x is equal to :

Inverse Trigonometric Functions PYQ — JEE Main 2002 · Shift 0 · Q76 with Solution

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Correct answer: A

Let $t=\sqrt{\cos\alpha}.$ Then the given equation becomes $\cot^{-1}(t)-\tan^{-1}(t)=x.$
Use the identity relating inverse cotangent and inverse tangent: $\cot^{-1}(t)=\frac{\pi}{2}-\tan^{-1}(t)$ (for principal values, with $t>0$ here since $t=\sqrt{\cos\alpha}$ ).

So,

=\frac{\pi}{2}-2\tan^{-1}(t).$$ 3. Now compute $\sin x$: $$\sin x=\sin\left(\frac{\pi}{2}-2\tan^{-1}(t)\right)=\cos\left(2\tan^{-1}(t)\right).$$ 4. Use the standard identity $$\cos(2\tan^{-1} t)=\frac{1-t^2}{1+t^2}.$$ Hence, $$\sin x=\frac{1-t^2}{1+t^2}.$$ Since $t^2=\cos\alpha$, we get $$\sin x=\frac{1-\cos\alpha}{1+\cos\alpha}.$$ 5. Simplify using half-angle identities: $$1-\cos\alpha=2\sin^2\frac{\alpha}{2}, \qquad 1+\cos\alpha=2\cos^2\frac{\alpha}{2}.$$ Therefore, $$\sin x=\frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)}=\tan^2\left(\frac{\alpha}{2}\right).$$ 6. Compare with options: - A: $\tan^2\left(\frac{\alpha}{2}\right)$ ✅ - B: $\cot^2\left(\frac{\alpha}{2}\right)$ ❌ - C: $\tan\alpha$ ❌ - D: $\cot\left(\frac{\alpha}{2}\right)$ ❌ Hence the correct option is **A**.

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View written solutionFree

Correct answer: A

Let $t=\sqrt{\cos\alpha}.$ Then the given equation becomes $\cot^{-1}(t)-\tan^{-1}(t)=x.$
Use the identity relating inverse cotangent and inverse tangent: $\cot^{-1}(t)=\frac{\pi}{2}-\tan^{-1}(t)$ (for principal values, with $t>0$ here since $t=\sqrt{\cos\alpha}$ ).

So,

=\frac{\pi}{2}-2\tan^{-1}(t).$$ 3. Now compute $\sin x$: $$\sin x=\sin\left(\frac{\pi}{2}-2\tan^{-1}(t)\right)=\cos\left(2\tan^{-1}(t)\right).$$ 4. Use the standard identity $$\cos(2\tan^{-1} t)=\frac{1-t^2}{1+t^2}.$$ Hence, $$\sin x=\frac{1-t^2}{1+t^2}.$$ Since $t^2=\cos\alpha$, we get $$\sin x=\frac{1-\cos\alpha}{1+\cos\alpha}.$$ 5. Simplify using half-angle identities: $$1-\cos\alpha=2\sin^2\frac{\alpha}{2}, \qquad 1+\cos\alpha=2\cos^2\frac{\alpha}{2}.$$ Therefore, $$\sin x=\frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)}=\tan^2\left(\frac{\alpha}{2}\right).$$ 6. Compare with options: - A: $\tan^2\left(\frac{\alpha}{2}\right)$ ✅ - B: $\cot^2\left(\frac{\alpha}{2}\right)$ ❌ - C: $\tan\alpha$ ❌ - D: $\cot\left(\frac{\alpha}{2}\right)$ ❌ Hence the correct option is **A**.