Question

Cosec theta + cot theta equal to p then prove that cos theta is equal to p square minus one upon p square + 1

Answer 1

Refer the attachment provided for answer.

Thnq

Answer 2

given

$\cosec \theta + \cot \theta = p$

Now RHS

$\frac{ {p}^{2} - 1}{ {p}^{2} + 1} \\ \\ = \frac{( {cosec \theta + \cot \theta})^{2} - 1}{( {cosec \theta + cot \theta})^{2} + 1} \\ \\ = \frac{2 {cot}^{2} \theta + 2cosec \theta cot \theta}{2 {cosec}^{2} \theta + 2cosec \theta cot \theta }$

$= \frac{2cot \theta(cot \theta + cosec \theta)}{2coec \theta(cosec \theta + cot \theta)} \\ \\ = \frac{cot \theta}{cosec \theta} \\ \\ = cos \theta$