KCET · Maths · Complex Number
If \( y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right) \), then \( \frac{d y}{d x} \) is equal to
- A \( 1 / 2 \)
- B \(\pi / 4\)
- C \( 0 \)
- D \( 1 \)
Answer & Solution
Correct Answer
(D) \( 1 \)
Step-by-step Solution
Detailed explanation
Given that, \( y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right) \)
\( =\tan ^{-1}\left(\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}\right)=\tan ^{-1}\left(\frac{\tan x+1}{1-\tan x}\right) \)
\( =\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+x\right)\right\}=\frac{I}{4}+x \)
\( \Rightarrow y=\frac{I}{4}+x \)
So, \( \frac{d y}{d x}=0+1=1 \)
\( =\tan ^{-1}\left(\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}\right)=\tan ^{-1}\left(\frac{\tan x+1}{1-\tan x}\right) \)
\( =\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+x\right)\right\}=\frac{I}{4}+x \)
\( \Rightarrow y=\frac{I}{4}+x \)
So, \( \frac{d y}{d x}=0+1=1 \)
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