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Simplify the following trigonometric equation:

$\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$

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\sec A cosec A + 1

\tan A + \cot A

\sec A + A

si n A \cos A + 1

answer is A.

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Detailed Solution

Given expression is,

\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}

= \frac{(\frac{1}{\tan A})}{1 - (\frac{1}{\tan A})} + \frac{1}{1 - \tan A}

(since,

\cot θ = \frac{1}{\tan θ}

)

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

= \frac{1 - \tan^{3} A}{\tan A (1 - \tan A)}

= \frac{(1 - \tan A) (1 + \tan A + \tan^{2} A)}{\tan A (1 - \tan A)} (since, a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2}))

= \frac{\sec^{2} A + \tan A}{\tan A} (since, 1 + \tan^{2} A = \sec^{2} A)

= 1 + \frac{\sec^{2} A}{\tan A}

= 1 + \frac{1}{\cos A \sin A}

= 1 + \sec A \csc A

(since,

\cos θ = \frac{1}{\sec θ}

and

\sin = \frac{1}{\csc θ}

)
So,

\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \sec A \csc A

Correct option is 1.

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