Question

If $\cos \theta =\frac{2t}{1+t^2}$, then find the value of $\tan \theta$.

Answer 1

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-{\left(\frac{2t}{1+t^2}\right)}^2}$
$\Rightarrow \sqrt{\frac{(1+t^2)-4t^2}{{(1+t^2)}^2}}=\sqrt{\frac{{(1-t^2)}^2}{1+t^2}}=\frac{1-t^2}{1+t^2}$
$\therefore \tan \theta =\frac{1-t^2}{1+t}+\frac{2t}{1+t^2}=\frac{1-t^2}{2t}$