JEE Advanced 2024 · Paper 2 · Mathematics Q1

MathematicsSingle correct+3 / -1Question 1 of 51

Considering only the principal values of the inverse trigonometric functions, the value of

$\tan\left(\sin^{-1}\left(\frac{3}{5}\right) - 2\cos^{-1}\left(\frac{2}{\sqrt5}\right)\right)$

\dfrac{7}{24}

\dfrac{-7}{24}

\dfrac{-5}{24}

\dfrac{5}{24}

Answer and worked solution

Official answer: B

Let $A=\sin^{-1}\frac35$ and $B=\cos^{-1}\frac{2}{\sqrt5}$ , both in $(0,\frac\pi2)$ .

$\tan A=\frac34$ .
$\cos B=\frac{2}{\sqrt5}$ gives $\sin B=\frac{1}{\sqrt5}$ , so $\tan B=\frac12$ and $\tan 2B=\dfrac{2\cdot\frac12}{1-\frac14}=\frac43$ .

$\tan(A-2B)=\frac{\frac34-\frac43}{1+\frac34\cdot\frac43}=\frac{-\frac{7}{12}}{2}=-\frac{7}{24}.$

JEE Advanced 2024 · Paper 2, Mathematics question 1