MHT CET · Maths · Inverse Trigonometric Functions
If \(2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)\), then the value of \(x\) is
- A \(\frac{\pi^c}{6}\)
- B \(\frac{\pi^c}{4}\)
- C \(\frac{\pi^c}{3}\)
- D \(\frac{\pi^\epsilon}{12}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi^c}{4}\)
Step-by-step Solution
Detailed explanation
We have \(2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)\)
\(
\begin{aligned}
& \therefore \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \\
& \therefore \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \Rightarrow \sin x \cos x=\sin ^2 x \\
& \therefore \sin x(\sin x-\cos x)=0 \Rightarrow \sin x=0 \text { or } \tan x=1 \\
& \therefore x=0 \text { or } x=\frac{\pi}{4}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \\
& \therefore \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \Rightarrow \sin x \cos x=\sin ^2 x \\
& \therefore \sin x(\sin x-\cos x)=0 \Rightarrow \sin x=0 \text { or } \tan x=1 \\
& \therefore x=0 \text { or } x=\frac{\pi}{4}
\end{aligned}
\)
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