^{-1}({ α})-^{-1}({ α})=x, then x= [AIEEE 2002]

Correct Answer is : tan⁡2(α2)²≤ft(α/2)tan2(2α)

Correct Answer is : tan⁡2(α2)²≤ft(α/2)tan2(2α)

Test Index

Inverse Trigonometric Functions

Section: Mathematics

Question : 99 of 99

Marks: +1, -0

\cot^{-1}(\sqrt{\cos \alpha})-\tan^{-1}(\sqrt{\cos \alpha})=x

\sin x=

Solution:

Given that,

\cot^{-1}(\sqrt{\cos x})-\tan^{-1}(\cos x)=x

.........(1)

We know,

\cot^{-1} x+\tan^{-1} x=\frac{\pi}{2}

\therefore \cot^{-1}(\sqrt{\cos x})+\tan^{-1}(\sqrt{\cos x})=\frac{\pi}{2}

.............(2)

Adding (1) and (2), we get,

2\cot^{-1}(\sqrt{\cos x})=x+\frac{\pi}{2}

\Rightarrow \cot^{-1}(\sqrt{\cos x})=\frac{x}{2}+\frac{\pi}{4}

\Rightarrow \sqrt{\cos x}=\cot\left(\frac{n}{2}+\frac{\pi}{4}\right)

\Rightarrow \sqrt{\cos x}=\frac{\cot\frac{x}{2}-1}{1+\cot\frac{x}{2}}

\Rightarrow \sqrt{\cos x}=\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}

Squaring both sides we get,

\Rightarrow \cos x=\frac{1-2\sin\frac{x}{2}\cos\frac{x}{2}}{1+2\sin\frac{x}{2}\cos\frac{x}{2}}

\Rightarrow \cos x=\frac{1-\sin x}{1+\sin x}

\Rightarrow \frac{1-\tan^2\frac{z}{2}}{1+\tan^2\frac{z}{2}}=\frac{1-\sin x}{1+\sin x}

Applying compounds and dividendo rule,

\Rightarrow \frac{2\sin x}{2}=\frac{2\tan^2\frac{x}{2}}{2}

\Rightarrow \sin x=\tan^2\frac{x}{2}

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