Question

Prove that (sec theta-cos theta) ( cot theta+tan theta) = tan theta ×sec theta​

Answer 1

Given:

$(\sec \theta \cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta$

To prove:

The given statement.

Proof:

We have,

$(\sec \theta -\cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta$

Taking LHS, we get

$LHS=(\sec \theta -\cos \theta)(\cot \theta+\tan \theta)$

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

$LHS=(\dfrac{1-\cos^2 \theta }{\cos \theta })(\dfrac{\cos^2 \theta+\sin^2 \theta}{\sin \theta\cos \theta})$

$LHS=(\dfrac{\sin^2 \theta }{\cos \theta })(\dfrac{1}{\sin \theta\cos \theta})$          $[\because \cos^2 \theta+\sin^2 \theta=1]$

$LHS=(\dfrac{\sin \theta }{\cos \theta })(\dfrac{1}{\cos \theta})$

$LHS=\tan \theta \sec \theta$

$LHS=RHS$

Hence proved.