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Question of 175

Q.Find the derivative of —

(i) sin⁡x+cos⁡xsin⁡x−cos⁡x\frac{\sin x + \cos x}{\sin x - \cos x}
(ii) (x+sec⁡x)(x−tan⁡x)(x + \sec x)(x - \tan x)
Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2026Subjective· 4mImportance★★★★★
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(i) The derivative is −21−sin⁡2x\dfrac{-2}{1-\sin 2x}. (ii) The derivative is (sec⁡xtan⁡x+1)(x−tan⁡x)+(x+sec⁡x)(1−sec⁡2x)(\sec x\tan x+1)(x-\tan x)+(x+\sec x)(1-\sec^2x).

(i) Let u=sin⁡x+cos⁡xu=\sin x+\cos x, v=sin⁡x−cos⁡xv=\sin x-\cos x. Then u′=cos⁡x−sin⁡xu'=\cos x-\sin x, v′=cos⁡x+sin⁡xv'=\cos x+\sin x.

By the quotient rule, ddx(uv)=u′v−uv′v2\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{u'v-uv'}{v^2}.

u′v=(cos⁡x−sin⁡x)(sin⁡x−cos⁡x)=−(cos⁡x−sin⁡x)2u'v = (\cos x-\sin x)(\sin x-\cos x) = -(\cos x-\sin x)^2. Expanding directly: (cos⁡x−sin⁡x)(sin⁡x−cos⁡x)=sin⁡2x−1(\cos x-\sin x)(\sin x-\cos x) = \sin2x - 1 (using 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x=\sin2x and sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1).

uv′=(sin⁡x+cos⁡x)(sin⁡x+cos⁡x)=(sin⁡x+cos⁡x)2=1+sin⁡2xuv' = (\sin x+\cos x)(\sin x+\cos x) = (\sin x+\cos x)^2 = 1+\sin2x.

u′v−uv′=(sin⁡2x−1)−(1+sin⁡2x)=−2u'v-uv' = (\sin2x-1)-(1+\sin2x) = -2.

v2=(sin⁡x−cos⁡x)2=1−sin⁡2xv^2 = (\sin x-\cos x)^2 = 1-\sin2x.

So the derivative is −21−sin⁡2x\dfrac{-2}{1-\sin2x}.

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