Question

Show that √(1-cos A/1+cos A) =cosec A - cot A​

Answer 1

Hi there!

$\sqrt{\frac{1-cosA}{1+cosA}} =$

We can begin by multiplying by its conjugate:

$\sqrt{\frac{1-cosA}{1+cosA}} * \sqrt{\frac{1+cosA}{1+cosA}} = \\\\\sqrt{\frac{(1-cosA)(1 + cosA)}{(1+cosA)(1 + cosA)}} =$

Simplify using the identity:

$1 - cos^2A = sin^2A$

$\sqrt{\frac{(1-cos^2A)}{(1+cosA)^2}} =\\\\\sqrt{\frac{(sin^2A)}{(1+cosA)^2}} =$

Take the square root of the expression:

${\frac{sinA}{1+cosA} =$

Multiply again by the conjugate to get a SINGLE term in the denominator:

${\frac{sinA}{1+cosA} * {\frac{1-cosA}{1-cosA} =$

Simplify:

${\frac{sinA(1-cosA)}{1-cos^2A} =$

Use the above trig identity one more:

${\frac{sinA(1-cosA)}{sin^2A} =$

Cancel out sinA:

${\frac{(1-cosA)}{sinA} =$

Split the fraction into two:

${\frac{1}{sinA} - \frac{cosA}{sinA} =$

Recall:

$1/sinA = cscA\\\\cosA/sinA = cotA$

Simplify:

$\frac{1}{sinA} + \frac{cosA}{sinA} = \boxed{cscA - cotA}$