Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

Let $y=log_e\left(\frac{a+bsinx}{a-bsinx}\right),$ then value of $\frac{dy}{dx}$ is :

Options:

$\frac{abcos x}{a^2-b^2sin^2x}$

$\frac{abcos x}{a^2+b^2sin^2x}$

$\frac{asin x}{a^2-b^2sin^2x}$

$\frac{2abcos x}{a^2-b^2sin^2x}$

Correct Answer:

$\frac{2abcos x}{a^2-b^2sin^2x}$

Explanation:

The correct answer is option (4) → $\frac{2ab\cos x}{a^2-b^2\sin^2x}$

$y=\log_e\left(\frac{a+b\sin x}{a-b\sin x}\right)$

$\frac{dy}{dx}=\frac{a-b\sin x}{a+b\sin x}\frac{(b\cos x(a-b\sin x)+b\cos x(a+b\sin x))}{(a-b\sin x)^2}$

$=\frac{2ab\cos x}{a^2-b^2\sin^2x}$