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Q.

$\frac{1 + \tan^{2} A}{1 + \cot^{2} A} =$

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a

\sec^{2} A

b

- 1

c

\cot^{2} A

d

\tan^{2} A

answer is D.

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Detailed Solution

The given expression is

\frac{1 + \tan^{2} A}{1 + \cot^{2} A}

.
We know the trigonometric identity

\sin^{2} A + \cos^{2} A = 1

\frac{1 + \tan^{2} A}{1 + \cot^{2} A}

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

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

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

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

= \tan^{2} A

So, the value of the given expression

\frac{1 + \tan^{2} A}{1 + \cot^{2} A}

is

\tan^{2} A .

Hence, the correct option is (4).

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1+tan2A1+cot2A=