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EXERCISE 1.4 · Q174

Q.Differentiate tan⁡−1x1−x2\tan^{-1}\dfrac{x}{\sqrt{1-x^2}} w.r.t. sec⁡−112x2−1\sec^{-1}\dfrac{1}{2x^2-1}.

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Let u=tan⁡−1x1−x2u=\tan^{-1}\dfrac{x}{\sqrt{1-x^2}}, v=sec⁡−112x2−1v=\sec^{-1}\dfrac{1}{2x^2-1}. Put x=sin⁡θx=\sin\theta.

Step 1. Since x1−x2=sin⁡θcos⁡θ=tan⁡θ\dfrac{x}{\sqrt{1-x^2}}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta,

u=tan⁡−1(tan⁡θ)=θ=sin⁡−1xu=\tan^{-1}(\tan\theta)=\theta=\sin^{-1}x

Step 2. Since 2x2−1=2sin⁡2θ−1=−cos⁡2θ2x^2-1=2\sin^2\theta-1=-\cos2\theta, we get 12x2−1=−sec⁡2θ\dfrac{1}{2x^2-1}=-\sec2\theta, and

v=sec⁡−1(−sec⁡2θ)=sec⁡−1[sec⁡(π−2θ)]=π−2θv=\sec^{-1}(-\sec2\theta)=\sec^{-1}[\sec(\pi-2\theta)]=\pi-2\theta …

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