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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

If 0 < α < 30^ , then find the value of _2( 8 (30^ -2 α)- 8 α+ (30^...

If 0<α<30∘0 < \alpha < 30^{\circ}, then find the value of log⁡2(8cos⁡(30∘−2α)−8cot⁡α+cot⁡(30∘−α)3)\log _2\left(\frac{8 \cos \left(30^{\circ}-2 \alpha\right)-\frac{8}{\cot \alpha+\cot \left(30^{\circ}-\alpha\right)}}{\sqrt{3}}\right).

Correct answer

2

Step-by-step solution

E=log⁡2(P3)∴P=8cos⁡(30−2α)−8sin⁡α⋅sin⁡(30−α)cos⁡αsin⁡(30−α)+cos⁡(30−α)sin⁡α=8cos⁡(30−2α)−4(cos⁡(30−2α)−cos⁡30∘)sin⁡30∘=8cos⁡(30−2α)−8cos⁡(30−2α)+43=43E=log⁡2(P3)=log⁡24=2\begin{aligned} E & =\log _2\left(\frac{P}{\sqrt{3}}\right) \\ \therefore \quad P & =8 \cos (30-2 \alpha)-\frac{8 \sin \alpha \cdot \sin (30-\alpha)}{\cos \alpha \sin (30-\alpha)+\cos (30-\alpha) \sin \alpha} \\ & =8 \cos (30-2 \alpha)-\frac{4\left(\cos (30-2 \alpha)-\cos 30^{\circ}\right)}{\sin 30^{\circ}} \\ & =8 \cos (30-2 \alpha)-8 \cos (30-2 \alpha)+4 \sqrt{3}=4 \sqrt{3} \\ E & =\log _2\left(\frac{P}{\sqrt{3}}\right)=\log _2 4=2\end{aligned}
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