Q59 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics

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

  • A
    $\tan \mathrm{A}+\cot \mathrm{A}$
  • B
    $\sec \mathrm{A}+\operatorname{cosec} \mathrm{A}$
  • C
    $\sin A \cos A+1$
  • Correct answer
    D
    $\sec A \operatorname{cosec} A+1$

Solution

Since, $\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}$ $\begin{aligned} & =\frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}+\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}} \\ & =\frac{\cos ^2 A}{\sin A(\cos A-\sin A)}+\frac{\sin ^2 A}{\cos A(\sin A-\cos A)} \\ & =\frac{(\cos A-\sin A)(1+\sin A \cos A)}{(\cos A-\sin A) \sin A \cdot \cos A} \\ & =\sec A \cdot \operatorname{cosec} A+1 \end{aligned}$

Hence, the correct answer is option: 'D'.

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