sin θ1+cos θis equal to


$\frac{\sin \theta}{1+\cos \theta}$ is equal to

(a) $\frac{1+\cos \theta}{\sin \theta}$

(b) $\frac{1-\cos \theta}{\cos \theta}$

(c) $\frac{1-\cos \theta}{\sin \theta}$

(d) $\frac{1-\sin \theta}{\cos \theta}$


The given expression is $\frac{\sin \theta}{1+\cos \theta}$.

Multiplying both the numerator and denominator under the root by $(1-\cos \theta)$, we have

$\frac{\sin \theta}{1+\cos \theta}$

$=\frac{\sin \theta(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)}$

$=\frac{\sin \theta(1-\cos \theta)}{1-\cos ^{2} \theta}$

$=\frac{\sin \theta(1-\cos \theta)}{\sin ^{2} \theta}$

$=\frac{1-\cos \theta}{\sin \theta}$

Therefore, the correct option is (c).



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