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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Let 0 < θ < (π)/(2) , such that ( ^2 2 θ+4 ^4 θ-4 ^2 θ ^2 θ)/(4-4...

Let 0<θ<π20 < \theta < \frac{\pi}{2}, such that sin⁡22θ+4sin⁡4θ−4sin⁡2θcos⁡2θ4−4sin⁡2θ−sin⁡22θ=7−437+43\frac{\sin ^2 2 \theta+4 \sin ^4 \theta-4 \sin ^2 \theta \cos ^2 \theta}{4-4 \sin ^2 \theta-\sin ^2 2 \theta}=\frac{7-4 \sqrt{3}}{7+4 \sqrt{3}}, then θ=πn\theta=\frac{\pi}{n} where nn is:

Correct answer

12

Step-by-step solution

4sin⁡4θ4cos⁡2θ−4sin⁡2θ⋅cos⁡2θ=sin⁡4θcos⁡2θ⋅cos⁡2θ=(tan⁡θ)4∴(tan⁡θ)4=(2−32+3)2=(2−3)4∴tan⁡θ=2−3⇒θ=π12\begin{array}{rlrl} & \frac{4 \sin ^4 \theta}{4 \cos ^2 \theta-4 \sin ^2 \theta \cdot \cos ^2 \theta} & =\frac{\sin ^4 \theta}{\cos ^2 \theta \cdot \cos ^2 \theta}=(\tan \theta)^4 \\ & \therefore & (\tan \theta)^4 & =\left(\frac{2-\sqrt{3}}{2+\sqrt{3}}\right)^2=(2-\sqrt{3})^4 \\ & \therefore & \tan \theta & =2-\sqrt{3} \Rightarrow \theta=\frac{\pi}{12}\end{array}
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