Question

Prove that cot theta/1-sintheta=Sec theta+tantheta

Answer 1

The given equation is:

\frac{cot{\theta}}{1-sin{\theta}}=sec{\theta}+tan{\theta}

1−sinθ

cotθ

=secθ+tanθ

Taking the LHS of the above equation, we get

\frac{cot{\theta}}{1-sin{\theta}}

1−sinθ

cotθ

Multiply and divide by (1+sin{\theta})(1+sinθ) , we get

=\frac{cos{\theta}(1+sin{\theta})}{1-sin^{2}{\theta}}

1−sin

2

θ

cosθ(1+sinθ)

=\frac{cos{\theta}+cos{\theta}sin{\theta}}{cos^{2}{\theta}}

cos

2

θ

cosθ+cosθsinθ

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

cosθ

1

+

cosθ

sinθ

=sec{\theta}+tan{\theta}secθ+tanθ

=RHS

Hence proved.

Answer 2

Hope this will help you