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Q)  Prove that \frac{(1 + \cot ^2 A)}{(1 + \tan ^ 2 A)} = \cot ^2 A

Ans: 

Method 1:

Step 1: Let’s start with LHS and put values of cot A and tan A:

LHS = \frac{(1 + \cot ^2 A)}{(1 + \tan ^ 2 A)}

= \frac{(1 + \frac{\cos ^2 A}{\sin ^2 A})}{(1 + \frac{\sin ^2 A}{\cos ^2 A})}

= \frac{(\frac{(\sin ^2 A + \cos ^2 A)}{\sin ^2 A})}{(\frac{(\cos ^2 A + \sin ^2 A)}{\cos ^2 A})}

Step 2: We know that sin 2  A + cos 2  A = 1

= \frac{(\frac{1}{\sin ^2 A})}{(\frac{1}{\cos ^2 A})}

= \frac{\cos ^2 A}{\sin ^2 A}

= \cot ^2A

Hence Proved !

Method 2:

Let’s start with LHS and applying trigonometric identities:

LHS = \frac{(1 + \cot ^2 A)}{(1 + \tan ^ 2 A)}

= \frac{(\cos ec^2 A)}{(\sec ^2 A)}

= \frac{(\frac{1}{\sin ^2 A})}{(\frac{1}{\cos ^2 A})}

= \frac{\cos ^2 A}{\sin ^2 A}

= \cot ^2A

Hence Proved !

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