Question

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

Answer 1

$\cos \theta =\frac{2t}{1+t^2}$

$\because 1+{\tan }^2\theta ={\sec }^2\theta =\frac{1}{{\cos }^2\theta }$

$1+{\tan }^2\theta =\frac{(1+t{)}^2}{(2t{)}^2}=\frac{1+t^2+2t}{4t^2}$

$=\frac{1}{4t^2}+\frac{1}{4}+\frac{2t}{4t^2}$

$=\frac{1}{4t^2}+\frac{1}{4}+\frac{1}{2t}$

${\tan }^2\theta =\frac{1}{4t^2}-\frac{3}{4}+\frac{1}{2t}$

$=\frac{1-3t^2+2t}{4t^2}=\frac{-3t^2+3t-t+1}{4t^2}$

$=\frac{3t(-t+1)+(-t+1)}{4t^2}$

$=\frac{(3t-1)(1-t)}{4t^2}$

$\tan \theta =\sqrt{\frac{(3t-1)(1-t)}{4t^2}}=\frac{\sqrt{(3t-1)(1-t)}}{2t}$.