Question

prove that√ 1 - cos A /1 + cos A =cosec A - cot A​

Answer 1

$\sqrt{ \frac{1 - cos \theta}{1 + cos \theta} } = cosec \theta - cot \theta$

Taking LHS

$\sqrt{ \frac{1 - cos \theta}{1 + cos \theta} }$

Rationalise It with 1-cosA

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

We know

sin²A+Cos²A=1

sin²A=1-cos²A

$\sqrt{ \frac{(1 - cos \theta) {}^{2} }{ sin {}^{2} \theta } } \\ \\ \sqrt{ (\frac{(1 - cos \theta) }{sin \theta} ) {}^{2} } \\ \\ (\frac{1 - cos \theta}{sin \theta} ) {}^{2 \times \frac{1}{2} } \\ \\ \frac{1 - cos \theta}{sin \theta} \\ \\ \frac{1}{sin \theta} - \frac{cos \theta}{sin \theta}$

We know

$\frac{1}{sin \theta} = cosec \theta \\ \\ \frac{cos \theta}{sin \theta} = cot \theta$

So,

$\frac{1}{sin \theta} - \frac{cos \theta}{sin \theta} \\ \\ cosec \theta - cot \theta$

Hence Proved

LHS =RHS