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← JEE Main 2022 · 25 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 19 of 90

JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 19

The value of tan⁡−1(cos⁡(15π4)−1sin⁡(π4))\tan^{-1}\left(\dfrac{\cos\left(\frac{15\pi}4\right)-1}{\sin\left(\frac\pi4\right)}\right) is equal to :

Choose one option
Answer and worked solution

Official answer: B

Simplify. cos⁡15π4=cos⁡(−π4)=12\cos\frac{15\pi}4=\cos\left(-\frac\pi4\right)=\frac1{\sqrt2}, so the argument is 12−112=1−2=−tan⁡π8\dfrac{\frac1{\sqrt2}-1}{\frac1{\sqrt2}}=1-\sqrt2=-\tan\frac\pi8.

Value. tan⁡−1(−tan⁡π8)=−π8\tan^{-1}\left(-\tan\frac\pi8\right)=-\frac\pi8.

(B)

Topics:Inverse trigonometric functions
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