Question

2tan inverse 1/3 +tan inverse 1/ 4 ​

Answer 1

First We Have To Simplify

$\huge 2 \tan^{ - 1} (x)$

Using FoRmuLa

$\huge \fbox{2 \tan^{ - 1} (x) = \tan^{ - 1} ( \frac{ 2x }{1 - {x}^{2} } )}$

By Putting The Value of x i FoRmuLa

$\huge {2 \tan^{ - 1} ( \frac{1}{3} ) = \tan^{ - 1} ( \frac{ 2 \times \frac{1}{3} }{1 - { (\frac{1}{3}) }^{2} } )}$

$\huge 2 \tan^{ - 1} ( \frac{1}{3} ) = \tan^{ - 1} ( \frac{ 3}{4 } )$

Now Let's Get Back To Question!!

We Have To Add It Using Formula

$\huge \fbox{\tan^{ - 1} (x) + \tan^{ - 1} (y)= \tan^{ - 1} ( \frac{ x + y }{1 - x \times y } )}$

Putting The Values of x and y

$\huge \tan^{ - 1} ( \frac{ 3}{4 }) +\tan^{ - 1} ( \frac{ 1}{4 })$

$\huge \fbox{\tan^{ - 1} (\frac{3}{4}) + \tan^{ - 1} (\frac{1 }{4} )= \tan^{ - 1} ( \frac{\frac{3}{4} + \frac{1 }{4} }{1 - \frac{3}{4} \times \frac{1 }{4} } )}$

So The Answer Is

$\huge \fbox{ \blue{\tan^{ - 1} (x) + \tan^{ - 1} (y)= \tan^{ - 1} ( \frac{16}{13 } )}}$