Verified Solution • 2025–26 Session moderate

QJEEM21Maf397: JEE Main 2021 Mathematics - Class 11 JEEM21Maf397 Solution: The sum of solutions of the equation , is :

The sum of solutions of the equation

cos⁡x1+sin⁡x=∣tan⁡2x∣{{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right| , x∈(−π2,π2)−{π4,−π4}x \in \left( { - {\pi \over 2},{\pi \over 2}} \right) - \left\{ {{\pi \over 4}, - {\pi \over 4}} \right\} is :

Multiple Choice Options

  • A −11π30- {{11\pi } \over {30}}
  • B π10{\pi \over {10}}
  • C −7π30- {{7\pi } \over {30}}
  • D −π15- {\pi \over {15}}

Step-by-Step Verified Solution

cos⁡x1+sin⁡x=∣tan⁡2x∣{{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right|

⇒cos⁡2x/2−sin⁡2x/2(cos⁡x/2+sin⁡x/2)2=∣tan⁡2x∣\Rightarrow {{{{\cos }^2}x/2 - {{\sin }^2}x/2} \over {(\cos x/2 + \sin x/2)^2}} = \left| {\tan 2x} \right|

⇒\Rightarrow cos⁡x2−sin⁡x2cos⁡x2+sin⁡x2=∣tan⁡2x∣{{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}} = \left| {\tan 2x} \right|

⇒\Rightarrow 1−tan⁡x21+tan⁡x2=∣tan⁡2x∣{{1 - \tan {x \over 2}} \over {1 + \tan {x \over 2}}} = \left| {\tan 2x} \right|

⇒\Rightarrow tan⁡π4−tan⁡x2tan⁡π4+tan⁡x2=∣tan⁡2x∣{{\tan {\pi \over 4} - \tan {x \over 2}} \over {\tan {\pi \over 4} + \tan {x \over 2}}} = \left| {\tan 2x} \right|

⇒tan⁡2(π4−π2)=tan⁡22x\Rightarrow {\tan ^2}\left( {{\pi \over 4} - {\pi \over 2}} \right) = {\tan ^2}2x

⇒2x=nπ±(π4−π2)\Rightarrow 2x = n\pi \pm \left( {{\pi \over 4} - {\pi \over 2}} \right)

⇒x=−3π10,−π6,π10\Rightarrow x = {{ - 3\pi } \over {10}},{{ - \pi } \over 6},{\pi \over {10}}

or sum =−11π6= {{ - 11\pi } \over 6} .

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