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EXERCISE 1.2 · Q94

Q.Differentiate the following w.r.t. xx: sin⁡−11−25x21+25x2\sin^{-1}\dfrac{1-25x^2}{1+25x^2}

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Put 5x=tan⁡θ5x=\tan\theta. Then 1−25x21+25x2=1−tan⁡2θ1+tan⁡2θ=cos⁡2θ=sin⁡(π2−2θ)\dfrac{1-25x^2}{1+25x^2}=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta=\sin\left(\dfrac\pi2-2\theta\right). So y=sin⁡−1[sin⁡(π2−2θ)]=π2−2θ=π2−2tan⁡−1(5x)y=\sin^{-1}\left[\sin\left(\dfrac\pi2-2\theta\right)\right]=\dfrac\pi2-2\theta=\dfrac\pi2-2\tan^{-1}(5x). Diffe …

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