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Question 113 of 143

Q.Find dydx\dfrac{dy}{dx} if tan⁡(x+y)+tan⁡(x−y)=1\tan(x+y) + \tan(x-y) = 1.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2018Subjective· 3mImportance★★★★★
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Implicit differentiation of tan⁡(x+y)+tan⁡(x−y)=1\tan(x+y)+\tan(x-y)=1 gives dydx=sec⁡2(x+y)+sec⁡2(x−y)sec⁡2(x−y)−sec⁡2(x+y)\dfrac{dy}{dx} = \dfrac{\sec^2(x+y)+\sec^2(x-y)}{\sec^2(x-y)-\sec^2(x+y)}.

Differentiate both sides of tan⁡(x+y)+tan⁡(x−y)=1\tan(x+y)+\tan(x-y)=1 with respect to xx, treating yy as a function of xx:

sec⁡2(x+y)⋅ddx(x+y)+sec⁡2(x−y)⋅ddx(x−y)=0\sec^2(x+y)\cdot\dfrac{d}{dx}(x+y) + \sec^2(x-y)\cdot\dfrac{d}{dx}(x-y) = 0

sec⁡2(x+y)(1+y′)+sec⁡2(x−y)(1−y′)=0\sec^2(x+y)(1+y') + \sec^2(x-y)(1-y') = 0

Let A=sec⁡2(x+y)A=\sec^2(x+y) and B=sec⁡2(x−y)B=\sec^2(x-y) for brevity:

A(1+y′)+B(1−y′)=0A(1+y')+B(1-y')=0

A+Ay′+B−By′=0A+Ay'+B-By'=0

y′(A−B)=−(A+B)y'(A-B) = -(A+B)

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