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← JEE Main 2022 · 26 Jun · Shift 1
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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 6

lim⁡x→12sin⁡(cos⁡−1x)−x1−tan⁡(cos⁡−1x)\displaystyle\lim_{x\to\frac1{\sqrt2}}\frac{\sin\left(\cos^{-1}x\right)-x}{1-\tan\left(\cos^{-1}x\right)} is equal to :

Choose one option
Answer and worked solution

Official answer: D

Substitute t=cos⁡−1x→π4t=\cos^{-1}x\to\frac\pi4, x=cos⁡tx=\cos t: the expression is sin⁡t−cos⁡t1−tan⁡t=sin⁡t−cos⁡t(cos⁡t−sin⁡t)/cos⁡t=−cos⁡t\dfrac{\sin t-\cos t}{1-\tan t}=\dfrac{\sin t-\cos t}{(\cos t-\sin t)/\cos t}=-\cos t.

Limit. −cos⁡π4=−12-\cos\frac\pi4=-\dfrac1{\sqrt2}.

(D)

Topics:LimitsInverse trigonometric functions
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