 Krishna
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Step 1: Take down the given equation

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

Step 2:  Take the L.H.S of the equation and think about the possible ways to prove R.H.S

NOTE:   \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}

Step 3: Rationalize the irrational denominator

EXPLANATION; You can't solve an equation that contains a fraction with an irrational denominator, which means that the denominator contains a term with a radical sign. This includes square, cube and higher roots. Getting rid of the radical sign is called rationalizing the denominator.

EXAMPLE: \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}

Divide and multiply by 1+\cos\theta

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

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

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

=     \frac{1 + \cos \theta}{\sin \theta}  (Since, \sin^2 \theta = 1 - \cos^2 \theta )

Step 3: Separate the denominator and simplify for R.H.S

EXAMPLE:   \frac{1}{\sin \theta} + \frac{ \cos \theta}{sin \theta}

\cosec \theta + \cot \theta

(since, \cosec \theta = \frac{1}{\sin \theta}

\cot \theta = \frac{\cos \theta}{\sin \theta}   )

Hence, proved