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Mathematics Question on Trigonometric Functions

The expression tan⁡A1−cot⁡A+cot⁡A1−tan⁡A\frac{\tan A}{1-\cot A} + \frac{\cot A}{1 - \tan A} can be written as

A

sec⁡A cosecA+1\sec A \, cosec A + 1

B

tan⁡A+cot⁡A\tan A + \cot A

C

sec⁡A+cosecA\sec A + cosec A

D

sin⁡A cos⁡A+1\sin A \, \cos A + 1

Answer

sec⁡A cosecA+1\sec A \, cosec A + 1

Explanation

Solution

tan⁡A1−cot⁡A+cot⁡A1−tan⁡A\frac{\tan A}{1-\cot A} + \frac{\cot A}{1 - \tan A}
=sin⁡2Acos⁡A(sin⁡A−cos⁡A)+cos⁡2Asin⁡A(cos⁡A−sin⁡A)= \frac{\sin^{2} A}{\cos A \left(\sin A - \cos A \right)}+ \frac{\cos^{2} A}{\sin A \left( \cos A - \sin A \right)}
=sin⁡3A−cos⁡3A(sin⁡A−cos⁡A)cos⁡Asin⁡A= \frac{\sin^{3} A - \cos^{3} A}{\left(\sin A - \cos A\right) \cos A \sin A }
=(sin⁡A−cos⁡A)(sin⁡2A+sin⁡Acos⁡A+cos⁡2A)(sin⁡A−cos⁡A)sin⁡Acos⁡A= \frac{\left(\sin A - \cos A\right)\left(\sin^{2} A + \sin A \cos A+ \cos^{2} A\right)}{\left(\sin A - \cos A\right) \sin A \cos A}
=1+sin⁡Acos⁡Asin⁡Acos⁡A=1+sec⁡AcosecA= \frac{1+\sin A \cos A}{\sin A \cos A} = 1 + \sec A cosecA