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← JEE Main 2022 · 26 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 19 of 90

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 19

If the inverse trignometric functions take principal values, then
cos⁡−1(310cos⁡(tan⁡−1(43))+25sin⁡(tan⁡−1(43)))\cos^{-1}\left(\frac3{10}\cos\left(\tan^{-1}\left(\frac43\right)\right)+\frac25\sin\left(\tan^{-1}\left(\frac43\right)\right)\right)
is equal to :

Choose one option
Answer and worked solution

Official answer: C

Angle tan⁡−143\tan^{-1}\frac43: cosine 35\frac35, sine 45\frac45.

Inside. 310⋅35+25⋅45=950+1650=12\frac3{10}\cdot\frac35+\frac25\cdot\frac45=\frac9{50}+\frac{16}{50}=\frac12, so the value is cos⁡−112=π3\cos^{-1}\frac12=\dfrac\pi3.

(C)

Topics:Inverse trigonometric functions
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