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CBSE Class 12 Math 2023 All Sets Solved Paper

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Question : 12 of 20
Marks: +1, -0
If tan⁡(x+yx−y)=k,\tan\left(\frac{x+y}{x-y}\right)=k, then dydx\frac{dy}{dx} is equal to :
Solution:  
tan⁡(x+yx−y)=k\tan\left(\frac{x+y}{x-y}\right)=k
⇒(x+y)(x−y)=tan⁡−1k\Rightarrow (x+y)(x-y)=\frac{\tan^{-1}}{k}
⇒(x−y)(1+y′)−(x+y)(1−y′)(x−y)2=0\Rightarrow \frac{(x-y)(1+y')-(x+y)(1-y')}{(x-y)^2}=0
⇒x−y+xy′−yy′−x−y+xy′+yy′=0\Rightarrow x-y+xy'-yy'-x-y+xy'+yy'=0
⇒−2y+2xy′=0\Rightarrow -2y+2xy'=0
∴dydx=yx\therefore \frac{dy}{dx}=\frac{y}{x}
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