Question

under root 1+cos theta/1-cos theta=cosec theta+cot theta

Answer 1

hey there hope it'll help ya

Answer 2

Answer:

To prove: \sqrt{\frac{1+cos{\theta}}{1-cos{\theta}}}=cosec{\theta}+cot{\theta}

1−cosθ

1+cosθ

=cosecθ+cotθ

Taking the LHS of the above equation,

\sqrt{\frac{1+cos{\theta}}{1-cos{\theta}}}=\sqrt{\frac{1+cos{\theta}}{1-cos{\theta}}{\times} \frac{1+cos{\theta}}{1+cos{\theta}}}

1−cosθ

1+cosθ

=

1−cosθ

1+cosθ

×

1+cosθ

1+cosθ

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

1−cos

2

θ

(1+cosθ)

2

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

sin

2

θ

(1+cosθ)

2

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

sinθ

1+cosθ

=cosec{\theta}+cot{\theta}cosecθ+cotθ =RHS

Hence proved.