tanA1−cotA+cotA1−tanA=

tanA1−cotA+cotA1−tanA=sinAcosA×sinAsinA−cosA+cosAsinA×cosAcosA−sinA=1sinA−cosAsin2AcosA−cos2AsinA=sin3A−cos3A(sinA−cosA)cosAsinA=sin2A+cos2A+sinAcosAcosAsinA=1+sinAcosAcosAsinA=secAcosceA+1

tanA1−cotA+cotA1−tanA=

tanA1−cotA+cotA1−tanA=sinAcosA×sinAsinA−cosA+cosAsinA×cosAcosA−sinA=1sinA−cosAsin2AcosA−cos2AsinA=sin3A−cos3A(sinA−cosA)cosAsinA=sin2A+cos2A+sinAcosAcosAsinA=1+sinAcosAcosAsinA=secAcosceA+1

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

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sec A cosec A + 1

sin A cos A + 1

tan A + cot A

sec A + cosec A

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = \frac{\sin A}{\cos A} \times \frac{\sin A}{\sin A - \cos A} + \frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A - \sin A} \\ = \frac{1}{\sin A - \cos A} [\frac{\sin^{2} A}{\cos A} - \frac{\cos^{2} A}{\sin A}] = \frac{\sin^{3} A - \cos^{3} A}{(\sin A - \cos A) \cos A \sin A} \\ = \frac{\sin^{2} A + \cos^{2} A + \sin A \cos A}{\cos A \sin A} = \frac{1 + \sin A \cos A}{\cos A \sin A} = \sec A \cos c e A + 1 \end{array}$