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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 24 of 61
Marks: +1, -0
tan⁡(π4+x)tan⁡(π4−x)\frac{\tan\left(\frac{\pi}{4}+x\right)}{\tan\left(\frac{\pi}{4}-x\right)} = (1+tan⁡x1−tan⁡x)2\left(\frac{1+\tan x}{1-\tan x}\right)^2
Solution:  
We have, L.H.S. = tan⁡(π4+x)tan⁡(π4−x)\frac{\tan\left(\frac{\pi}{4}+x\right)}{\tan\left(\frac{\pi}{4}-x\right)} = tan⁡π4+tan⁡x1−tan⁡π4⋅tan⁡xtan⁡π4−tan⁡x1+tan⁡π4⋅tan⁡x\frac{\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\cdot\tan x}}{\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\cdot\tan x}}
Since tan (A + B) = tan⁡A+tan⁡B1−tan⁡Atan⁡B\frac{\tan A+\tan B}{1-\tan A\tan B} and tan (A - B) = tan⁡A−tan⁡B1+tan⁡Atan⁡B\frac{\tan A-\tan B}{1+\tan A\tan B}
= 1+tan⁡x1−tan⁡x1−tan⁡x1+tan⁡x\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}} = 1+tan⁡x1−tan⁡x\frac{1+\tan x}{1-\tan x} × 1+tan⁡x1−tan⁡x\frac{1+\tan x}{1-\tan x} = (1+tan⁡x1−tan⁡x)2\left(\frac{1+\tan x}{1-\tan x}\right)^2 = R.H.S.
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