Question

Prove the following :

$\frac{\cos ({90}^{\circ }-\theta )\sec ({90}^{\circ }-\theta )\tan \theta }{\text{cosec}({90}^{\circ }-\theta )\sin ({90}^{\circ }-\theta )\cot ({90}^{\circ }-\theta )}+\frac{\tan ({90}^{\circ }-\theta )}{\cot \theta }=2$

Answer 1

To prove :

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

We know that,

$\cos (90-\theta )=\sin \theta$

$\sin (90-\theta )=\cos \theta$

$\tan (90-\theta )=\cot \theta$

$\cot (90-\theta )=\tan \theta$

$\sec (90-\theta )=\csc \theta$

$\csc (90-\theta )=\sec \theta$

Also,

$\sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}$ and $\csc \left(\theta \right)=\frac{1}{\sin \left(\theta \right)}$ ..... (1)

So, rewriting the given equation with these identites, we get

$\frac{\sin \theta \csc \theta \tan \theta }{\sec \theta \cos \theta \tan \theta }+\frac{\cot \theta }{\cot \theta }=2$

$\frac{\sin \theta \csc \theta }{\sec \theta \cos \theta }+1=2$

$\frac{\sin \theta \cos \theta }{\sin \theta \cos \theta }+1=2$ [From (1)]

$1+1=2$

$2=2$

Hence proved