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QJEEAdv09Ma860f: JEE Advanced 2009 Mathematics - Class 11 JEEAdv09Ma860f Solution: For the solution (s) of is (are)

For 0 &lt; \theta &lt; {\pi \over 2}, the solution (s) of </mo></mrow><munderover><modata−mjx−texclass="OP"movablelimits="false">∑</mo><mrowdata−mjx−texclass="ORD"><mi>m</mi><mo>=</mo><mn>1</mn></mrow><mn>6</mn></munderover><mrowdata−mjx−texclass="ORD"><mi>cos</mi><modata−mjx−texclass="NONE">⁡</mo><mi>e</mi><mi>c</mi><mstylescriptlevel="0"><mspacewidth="0.167em"></mspace></mstyle><mrowdata−mjx−texclass="INNER"><modata−mjx−texclass="OPEN">(</mo><mrowdata−mjx−texclass="ORD"><mi>θ</mi><mo>+</mo><mrowdata−mjx−texclass="ORD"><mfrac><mrowdata−mjx−texclass="ORD"><mrowdata−mjx−texclass="INNER"><modata−mjx−texclass="OPEN">(</mo><mrowdata−mjx−texclass="ORD"><mi>m</mi><mo>−</mo><mn>1</mn></mrow><modata−mjx−texclass="CLOSE">)</mo></mrow><mi>π</mi></mrow><mn>4</mn></mfrac></mrow></mrow><modata−mjx−texclass="CLOSE">)</mo></mrow><mstylescriptlevel="0"><mspacewidth="0.167em"></mspace></mstyle><mi>cos</mi><modata−mjx−texclass="NONE">⁡</mo><mi>e</mi><mi>c</mi><mstylescriptlevel="0"><mspacewidth="0.167em"></mspace></mstyle><mrowdata−mjx−texclass="INNER"><modata−mjx−texclass="OPEN">(</mo><mrowdata−mjx−texclass="ORD"><mi>θ</mi><mo>+</mo><mrowdata−mjx−texclass="ORD"><mfrac><mrowdata−mjx−texclass="ORD"><mi>m</mi><mi>π</mi></mrow><mn>4</mn></mfrac></mrow></mrow><modata−mjx−texclass="CLOSE">)</mo></mrow><mo>=</mo><mn>4</mn><msqrt><mn>2</mn></msqrt></mrow></math></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container></mo></mrow><munderover><mo data-mjx-texclass="OP" movablelimits="false">∑</mo><mrow data-mjx-texclass="ORD"><mi>m</mi><mo>=</mo><mn>1</mn></mrow><mn>6</mn></munderover><mrow data-mjx-texclass="ORD"><mi>cos</mi><mo data-mjx-texclass="NONE">⁡</mo><mi>e</mi><mi>c</mi><mstyle scriptlevel="0"><mspace width="0.167em"></mspace></mstyle><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mrow data-mjx-texclass="ORD"><mi>θ</mi><mo>+</mo><mrow data-mjx-texclass="ORD"><mfrac><mrow data-mjx-texclass="ORD"><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mrow data-mjx-texclass="ORD"><mi>m</mi><mo>−</mo><mn>1</mn></mrow><mo data-mjx-texclass="CLOSE">)</mo></mrow><mi>π</mi></mrow><mn>4</mn></mfrac></mrow></mrow><mo data-mjx-texclass="CLOSE">)</mo></mrow><mstyle scriptlevel="0"><mspace width="0.167em"></mspace></mstyle><mi>cos</mi><mo data-mjx-texclass="NONE">⁡</mo><mi>e</mi><mi>c</mi><mstyle scriptlevel="0"><mspace width="0.167em"></mspace></mstyle><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mrow data-mjx-texclass="ORD"><mi>θ</mi><mo>+</mo><mrow data-mjx-texclass="ORD"><mfrac><mrow data-mjx-texclass="ORD"><mi>m</mi><mi>π</mi></mrow><mn>4</mn></mfrac></mrow></mrow><mo data-mjx-texclass="CLOSE">)</mo></mrow><mo>=</mo><mn>4</mn><msqrt><mn>2</mn></msqrt></mrow></math></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container> is (are)

Multiple Choice Options

  • A  π4\,{\pi \over 4}
  • B  π6\,{\pi \over 6 }
  • C  π12\,{\pi \over 12}
  • D  5π12\,{5\pi \over 12}

Step-by-Step Verified Solution

Given solutions

1sin⁡(π/4)[sin⁡(θ+π/4−θ)sin⁡θ . sin⁡(θ+π/4)+sin⁡(θ+π/2−(θ+π/4))sin⁡(θ+π/4) . (θ+π/2) + ... + sin⁡((θ+3π/2)−(θ+5π/4))sin⁡(θ+3π/2) . sin⁡(θ+5π/4)]=42{1 \over {\sin (\pi /4)}}\left[ {{{\sin (\theta + \pi /4 - \theta )} \over {\sin \theta \,.\,\sin (\theta + \pi /4)}} + {{\sin (\theta + \pi /2 - (\theta + \pi /4))} \over {\sin (\theta + \pi /4)\,.\,(\theta + \pi /2)}}\, + \,...\, + \,{{\sin ((\theta + 3\pi /2) - (\theta + 5\pi /4))} \over {\sin (\theta + 3\pi /2)\,.\,\sin (\theta + 5\pi /4)}}} \right] = 4\sqrt 2

⇒2[cos⁡θ−cot⁡(θ+π/4)+cot⁡(θ+π/4)−cot⁡(θ+π/2) + ... + cot⁡(θ+5π/4)−cot⁡(θ+3π/2)]=42\Rightarrow \sqrt 2 [\cos \theta - \cot (\theta + \pi /4) + \cot (\theta + \pi /4) - \cot (\theta + \pi /2)\, + \,...\, + \,\cot (\theta + 5\pi /4) - \cot (\theta + 3\pi /2)] = 4\sqrt 2

⇒tan⁡θ+cot⁡θ=4⇒tan⁡θ=2±3\Rightarrow \tan \theta + \cot \theta = 4 \Rightarrow \tan \theta = 2 \pm \sqrt 3

⇒θ=π12\Rightarrow \theta = {\pi \over {12}} or 5π12{{5\pi } \over {12}}

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