Question:
Let \(a \in \left(0, \frac{\pi}{2}\right)\) be fixed. If the integral \(\int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha}\,dx = A(x)\cos 2\alpha + B(x)\sin 2\alpha + C\), where C is a constant of integration, then the functions A(x) and B(x) are respectively :
Let \(a \in \left(0, \frac{\pi}{2}\right)\) be fixed. If the integral \(\int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha}\,dx = A(x)\cos 2\alpha + B(x)\sin 2\alpha + C\), where C is a constant of integration, then the functions A(x) and B(x) are respectively :
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Write the integrand as \(\frac{\sin(x+\alpha)}{\sin(x-\alpha)}\), put \(t = x-\alpha\) and expand \(\sin(t+2\alpha)\).
Updated On: Nov 11, 2026
- \(x-\alpha\) and \(\log_e|\cos(x-\alpha)|\)
- \(x+\alpha\) and \(\log_e|\sin(x-\alpha)|\)
- \(x+\alpha\) and \(\log_e|\sin(x+\alpha)|\)
- \(x-\alpha\) and \(\log_e|\sin(x-\alpha)|\)
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The Correct Option is D
Solution and Explanation
\(\frac{\tan x+\tan\alpha}{\tan x-\tan\alpha} = \frac{\sin x\cos\alpha+\cos x\sin\alpha}{\sin x\cos\alpha-\cos x\sin\alpha} = \frac{\sin(x+\alpha)}{\sin(x-\alpha)}\).
Let \(t = x-\alpha\), so \(x+\alpha = t+2\alpha\) and \(dx = dt\).
\(\frac{\sin(t+2\alpha)}{\sin t} = \cos 2\alpha + \sin 2\alpha\cot t\).
\(\int = t\cos 2\alpha + \sin 2\alpha\,\log_e|\sin t| + C = (x-\alpha)\cos 2\alpha + \log_e|\sin(x-\alpha)|\sin 2\alpha + C\).
Hence \(A(x) = x-\alpha\), \(B(x) = \log_e|\sin(x-\alpha)|\).
Answer: (4) \(x-\alpha\) and \(\log_e|\sin(x-\alpha)|\)
Let \(t = x-\alpha\), so \(x+\alpha = t+2\alpha\) and \(dx = dt\).
\(\frac{\sin(t+2\alpha)}{\sin t} = \cos 2\alpha + \sin 2\alpha\cot t\).
\(\int = t\cos 2\alpha + \sin 2\alpha\,\log_e|\sin t| + C = (x-\alpha)\cos 2\alpha + \log_e|\sin(x-\alpha)|\sin 2\alpha + C\).
Hence \(A(x) = x-\alpha\), \(B(x) = \log_e|\sin(x-\alpha)|\).
Answer: (4) \(x-\alpha\) and \(\log_e|\sin(x-\alpha)|\)
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