Question

$\frac{\cos ({90}^o-\theta )\sec ({90}^o-\theta )\tan \theta }{\csc ({90}^o-\theta )\sin ({90}^o-\theta )\cot ({90}^o-\theta )}+\frac{\tan ({90}^o-\theta )}{\cot \theta }=2$

• A. True
• B. False

Answer 1

The correct option is A True

LHS :

$\frac{\cos ({90}^o-\theta )\sec ({90}^o-\theta )\tan \theta }{\csc ({90}^o-\theta )\sin ({90}^o-\theta )\cot ({90}^o-\theta )}+\frac{\tan ({90}^o-\theta )}{\cot \theta }$

$=\frac{\sin \theta \csc \theta \tan \theta }{\sec \theta \cos \theta \tan \theta }+\frac{\cot \theta }{\cot \theta }$

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

$=1+1=2$ = RHS