Question

If $0^{\circ }<\theta <{90}^{\circ }$ and $\frac{\sin \theta }{1-\cos \theta }+\frac{\sin \theta }{1+\cos \theta }=4,$ then the value of $\theta$ is

• A. 30${}^{\circ }$
• B. 45${}^{\circ }$
• C. 60${}^{\circ }$
• D. none

Answer 1

The correct option is A 30${}^{\circ }$

Given, $\frac{sin\theta }{1-cos\theta }+\frac{sin\theta }{1+cos\theta }=4$
Taking LCM and cross multiplying we get
$\frac{sin\theta (1+cos\theta +1-cos\theta )}{(1-cos\theta )(1+cos\theta )}=4$
$=>\frac{sin\theta (1+1)}{(1-co{s}^2\theta )}=4$
$=>\frac{2sin\theta )}{\left(si{n}^2\theta \right)}=4$ (Since $1-co{s}^2\theta =si{n}^2\theta )$
$=>sin\theta =\frac{1}{2}$
$=>\theta ={30}^0$