Question

Prove that under root 1- cos Theta upon 1+ cos theta =cosec theta - cot theta

Answer 1

Answer:

Step-by-step explanation:

$\large\rm{ LHS=\sqrt{\dfrac{1-cos \theta}{1+cos \theta}}}$

$\large\rm{ =\sqrt{\dfrac{1-cos \theta}{1+cos \theta}}×\sqrt{\dfrac{1-cos \theta}{1-cos \theta}}}$

$\large\rm{= \dfrac{\sqrt{(1-cos \theta)^2}}{\sqrt{1-cos^2 \theta}}}$

$\large\rm{ = \dfrac{1-cos \theta}{\sqrt{sin^2 \theta}}}$

$\large\rm{= \dfrac{1-cos \theta}{sin \theta}}$

$\large\rm{= \dfrac{1}{sin \theta} - \dfrac{cos \theta}{sin \theta}}$

$\large\rm{= cosec \theta - cot \theta}$

$\large\rm{ LHS=RHS}$