Find the value of sin ≤ft( 2 tan^-1 1/4 ) + cos ≤ft( tan^-1 2√2 ).

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 $\tan^{-1} 2\sqrt{2} = \phi$

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

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

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