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=((1)/(\cos \theta )-\cos \theta)((\cos \theta)/(\sin \theta)+(\sin \theta)/(\cos \theta))`

`LHS=((1-\cos^2 \theta )/(\cos \theta ))((\cos^2 \theta+\sin^2 \theta)/(\sin \theta\cos \theta))`

`LHS=((\sin^2 \theta )/(\cos \theta ))((1)/(\sin \theta\cos \theta))`
`[\because \cos^2 \theta+\sin^2 \theta=1]`

`LHS=((\sin \theta )/(\cos \theta ))((1)/(\cos \theta))`

`LHS=\tan \theta \sec \theta`

`LHS=RHS`

Hence proved.