Target Exam

CUET

Subject

Section B1

Chapter

Inverse Trigonometric Functions

Question:

Find the value of $\sin \left( 2 \tan^{-1} \frac{1}{4} \right) + \cos \left( \tan^{-1} 2\sqrt{2} \right)$.

Options:

$\frac{41}{51}$

$\frac{31}{51}$

$\frac{45}{51}$

$\frac{17}{31}$

Correct Answer:

$\frac{41}{51}$

Explanation:

The correct answer is Option (1) → $\frac{41}{51}$ ##

$\sin \left( 2 \tan^{-1} \frac{1}{4} \right) + \cos \left( \tan^{-1} 2\sqrt{2} \right)$

Let's first evaluate $\sin\left( 2 \tan^{-1} \frac{1}{4} \right)$:

Put $\tan^{-1} \frac{1}{4} = \theta$

$\Rightarrow \tan \theta = \frac{1}{4}$

$\text{Now, } \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2 \times \frac{1}{4}}{1 + \left( \frac{1}{4} \right)^2} = \frac{8}{17}$

To evaluate $\cos(\tan^{-1} 2\sqrt{2})$, put $\tan^{-1} 2\sqrt{2} = \phi$

$\Rightarrow \tan \phi = 2\sqrt{2}$

$\Rightarrow \cos \phi = \frac{1}{3}$

$\sin \left( 2 \tan^{-1} \frac{1}{4} \right) + \cos \left( \tan^{-1} 2\sqrt{2} \right) = \frac{8}{17} + \frac{1}{3} = \frac{41}{51}$