# Thread: Using calculus to deduce a trigonometric identity.

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

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

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

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

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

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

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

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