Verified Solution • 2025–26 Session moderate

QJEEM13Ma87c6: JEE Main 2013 Mathematics - Class 11 JEEM13Ma87c6 Solution: The expression can be written as:

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

Multiple Choice Options

  • A sin⁡A cos⁡A+1\sin {\rm A}\,\cos {\rm A} + 1
  • B  sec⁡A cos⁡ecA+1\,\sec {\rm A}\,\cos ec{\rm A} + 1
  • C tan⁡A+cot⁡A\tan {\rm A} + \cot {\rm A}
  • D sec⁡A+cos⁡ecA\sec {\rm A} + \cos ec{\rm A}

Step-by-Step Verified Solution

Given expression can be written as

sin⁡Acos⁡A×sin Asin⁡A−cos⁡A+cos⁡Asin⁡A×cos⁡Acos⁡A−sin A{{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}

(As tan⁡A=sin⁡Acos⁡A\tan A = {{\sin A} \over {\cos A}} and cot⁡A=cos⁡Asin⁡A\cot A = {{\cos A} \over {\sin A}} )

=1sin⁡A−cos⁡A{sin⁡3A−cos⁡3Acos⁡Asin⁡A}= {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}

=sin⁡2A+sin⁡Acos⁡A+cos⁡2 Asin⁡Acos⁡A= {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}

=1+sec⁡ Acosec A= 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A

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