Find the value of $\sin \left( 2 \tan^{-1} \frac{1}{4} \right) + \cos \left( \tan^{-1} 2\sqrt{2} \right)$. |
$\frac{41}{51}$ $\frac{31}{51}$ $\frac{45}{51}$ $\frac{17}{31}$ |
$\frac{41}{51}$ |
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 $\tan^{-1} 2\sqrt{2} = \phi$ $\Rightarrow \tan \phi = 2\sqrt{2}$ $\Rightarrow \cos \phi = \frac{1}{3}$ $\sin \left( 2 \tan^{-1} \frac{1}{4} \right) + \cos \left( \tan^{-1} 2\sqrt{2} \right) = \frac{8}{17} + \frac{1}{3} = \frac{41}{51}$ |