Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

The value of $sin^{-1} \left(\frac{8}{17}\right) + sin^{-1} \left(\frac{3}{5}\right)$ is :

Options:

$sin^{-1}\left(\frac{77}{85}\right)$

$cos^{-1}\left(\frac{77}{85}\right)$

$tan^{-1}\left(\frac{77}{85}\right)$

$cot^{-1}\left(\frac{77}{85}\right)$

Correct Answer:

$sin^{-1}\left(\frac{77}{85}\right)$

Explanation:

The correct answer is Option (1) → $\sin^{-1}\left(\frac{77}{85}\right)$

$\sin^{-1} \left(\frac{8}{17}\right) + \sin^{-1} \left(\frac{3}{5}\right)=y$

$A=\sin^{-1}\frac{8}{17}$, $B=\sin^{-1}\frac{3}{5}$

$\sin A=\frac{8}{17}$, $\sin B=\frac{3}{5}$

Using $\cos θ=\sqrt{1-\sin^2θ}$

$\cos A=\frac{15}{17}$, $\cos B=\frac{4}{5}$

$A+B=\sin^{-1}(\sin(A+B))$

$=\sin^{-1}(\sin A\cos B+\sin B\cos A)$

$=\sin^{-1}\left(\frac{77}{85}\right)$