Given:
[tex](\sec \theta \cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta[/tex]
To prove:
The given statement.
Proof:
We have,
[tex](\sec \theta -\cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta[/tex]
Taking LHS, we get
[tex]LHS=(\sec \theta -\cos \theta)(\cot \theta+\tan \theta)[/tex]
[tex]LHS=(\dfrac{1}{\cos \theta }-\cos \theta)(\dfrac{\cos \theta}{\sin \theta}+\dfrac{\sin \theta}{\cos \theta})[/tex]
[tex]LHS=(\dfrac{1-\cos^2 \theta }{\cos \theta })(\dfrac{\cos^2 \theta+\sin^2 \theta}{\sin \theta\cos \theta})[/tex]
[tex]LHS=(\dfrac{\sin^2 \theta }{\cos \theta })(\dfrac{1}{\sin \theta\cos \theta})[/tex] [tex][\because \cos^2 \theta+\sin^2 \theta=1][/tex]
[tex]LHS=(\dfrac{\sin \theta }{\cos \theta })(\dfrac{1}{\cos \theta})[/tex]
[tex]LHS=\tan \theta \sec \theta[/tex]
[tex]LHS=RHS[/tex]
Hence proved.
