# tan(2tan inverseX + cot inverseX) at x=2. I couldnt understand the X=2 thing

Harish
82 Points
4 years ago
Hello Aashi !

The question is that we have to find the value of tan( 2tan-1x + cot-1x ) at x = 2.

So, we can rewrite the question as :
tan( tan-1x + tan-1x + cot-1x )
Since we know that tan-1x + cot-1x =  $\frac{\pi }{2}$

Therefore, the question now becomes :
tan(  $\frac{\pi }{2}$   + tan-1x )

If you see this looks like tan( a + b ) formula , where a = $\frac{\pi }{2}$ and b = tan-1x
Therefore, solving using the formula above we will get,

$\frac{tan(\frac{\pi }{2}) + tan(tan^{-1}x)}{1- tan(\frac{\pi }{2})tan(tan^{-1}x)}$

$\frac{\infty + x }{1 - (\infty )(x)}$

= 0