Question

If $\tan \theta +\sec \theta =4$, then value of: $\sin \theta \cos \theta$

Answer 1

$\tan \theta +\sec \theta =4$ ....(i)
We know that ${\sec }^2\theta -{\tan }^2\theta =1$ ....(ii)
Dividing (ii) by (i) we have
$\frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta +\tan \theta }=\frac{1}{4}$ i.e. $\sec \theta -\tan \theta =\frac{1}{4}$
$\therefore$ On addition of (i) and (ii) gives
$2\sec \theta =4+\frac{1}{4}=\frac{17}{4}$
or $\sec \theta =\frac{17}{8}$ i.e. $\cos \theta =\frac{8}{17}$
Now $\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{-\frac{64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17}$
$\therefore \sin \theta \cos \theta =\frac{8}{17}\times \frac{15}{17}=\frac{120}{189}$