# Thread: Using calculus to deduce a trigonometric identity.

1. ## Using calculus to deduce a trigonometric identity.

$g(x)=tan^{-1} x+ tan^{-1} (\frac{1}{x})$. Find $g'(x)$ and hence deduce that $tan^{-1} x + tan^{-1} (\frac{1}{x}) = \frac{\pi}{2}$

2. Originally Posted by nerdzor
$g(x)=tan^{-1} x+ tan^{-1} (\frac{1}{x})$. Find $g'(x)$ and hence deduce that $tan^{-1} x + tan^{-1} (\frac{1}{x}) = \frac{\pi}{2}$
Can you get g'(x)? If not, where do you get stuck? If you can get it, please post the answer you got (and show your working).

3. $
g(x)=tan^{-1} x+ tan^{-1} (\frac{1}{x})
$

$
g'(x)=\frac{1}{1+x^2}-\frac{1}{1+\frac{1}{x^2}}(\frac{-1}{x^2})=0
$

Since g'(x)=0,therefore g(x) must be a constant.

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

Now,you should have mentioned that $x>0$.

Otherwise if $x<0$ then $g(x)=-\frac{\pi}{2}$