Question

Plzz help meh i havent prepared ittt plzz friends

Answer 1

Question

$\rm \: the \: value \: of \: ( \tan {}^{ - 1} 2 + \tan {}^{ - 1} 3) \: \: is$

Solution:-

Property of inverse trigonometry

$\rm \tan {}^{ - 1} \: x + \tan {}^{ - 1} y =$

1) if xy < 1

$\rm \tan {}^{ - 1} ( \frac{x + y}{1 - xy} )$

2) if x >0 , y>0 and xy>1

$\rm \: \pi + \tan {}^{ - 1} ( \frac{x + y}{1 - xy} )$

3) if x<0 , y <0 and xy>1

$\rm \: - \pi + \tan {}^{ - 1} ( \frac{x + y}{1 - xy} )$

Using 2 nd property we get

$\rm \: \pi + \tan {}^{ - 1} ( \frac{x + y}{1 - xy} )$

$\to \rm \: ( \tan {}^{ - 1} 2 + \tan {}^{ - 1} 3)$

So x = 2 and y = 3 , we get

$\rm \: \pi + \tan {}^{ - 1} ( \frac{2 + 3}{1 - 2 \times 3} )$

$\rm \: \pi + \tan {}^{ - 1} ( \frac{5}{1 - 6} )$

$\rm \: \pi + \tan {}^{ - 1} ( \frac{ - 5}{5} )$

$\rm \: \pi + \tan {}^{ - 1} ( - 1 )$

Using property we get

$\to \rm \: \tan {}^{ - 1} ( - x) = - \tan {}^{ - 1} (x)$

we get

$\rm \: \pi - \tan {}^{ - 1} ( 1 )$

So,

$\rm \to \: \tan {}^{ - 1} 1 = \frac{\pi}{4}$

We get

$\to \rm \pi - \frac{\pi}{4}$

$\rm \to \: \frac{4\pi - \pi}{4}$

$\rm \: \to \: \frac{3\pi}{4}$

Answer:-

$\rm \implies \: \frac{3\pi}{4}$