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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

A+2 2 A+4 4 A+8 Cot 8 A=

tan⁡A+2tan⁡2A+4tan⁡4A+8Cot⁡8A=\tan A+2 \tan 2 A+4 \tan 4 A+8 \operatorname{Cot} 8 A=

Options

  1. Atan⁡A\tan A
  2. BCot⁡A\operatorname{Cot} A
  3. Ctan⁡2 A\tan 2 \mathrm{~A}
  4. DCot⁡2 A\operatorname{Cot} 2 \mathrm{~A}

Correct answer

B. Cot⁡A\operatorname{Cot} A

Step-by-step solution

(C) =tan⁡A+2tan⁡2 A+4tan⁡4 A+8×1−tan⁡24 A2tan⁡4 A=tan⁡A+2tan⁡2 A+4tan⁡4 A+4(1−tan⁡24 A)tan⁡4 A=tan⁡A+2tan⁡2 A+4tan⁡24 A+4−4tan⁡24 Atan⁡4 A=tan⁡A+2tan⁡2 A+4tan⁡4 A=tan⁡A+2tan⁡2 A+4cot⁡4 A=tan⁡A+2tan⁡2 A+4×1−tan⁡22 A2tan⁡2 A=tan⁡A+2tan⁡2 A+2(1−tan⁡22 A)tan⁡2 A=tan⁡A+2tan⁡22 A+2−2tan⁡22 Atan⁡2 A=tan⁡A+2tan⁡2 A=tan⁡A+2cot⁡2 A=2(1−tan⁡2 A)2tan⁡A=tan⁡A+1−tan⁡2 Atan⁡A=tan⁡2 A+1−tan⁡2 Atan⁡A=1tan⁡A=cot⁡A=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \tan 4 \mathrm{~A}+8 \times \frac{1-\tan ^{2} 4 \mathrm{~A}}{2 \tan 4 \mathrm{~A}}=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \tan 4 \mathrm{~A}+\frac{4\left(1-\tan ^{2} 4 \mathrm{~A}\right)}{\tan 4 \mathrm{~A}}=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+\frac{4 \tan ^{2} 4 \mathrm{~A}+4-4 \tan ^{2} 4 \mathrm{~A}}{\tan 4 \mathrm{~A}}=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+\frac{4}{\tan 4 \mathrm{~A}}=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \cot 4 \mathrm{~A}=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \times \frac{1-\tan ^{2} 2 \mathrm{~A}}{2 \tan 2 \mathrm{~A}}=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+\frac{2\left(1-\tan ^{2} 2 \mathrm{~A}\right)}{\tan 2 \mathrm{~A}}=\tan \mathrm{A}+\frac{2 \tan ^{2} 2 \mathrm{~A}+2-2 \tan ^{2} 2 \mathrm{~A}}{\tan 2 \mathrm{~A}}=\tan \mathrm{A}+\frac{2}{\tan 2 \mathrm{~A}}=\tan \mathrm{A}+2 \cot 2 \mathrm{~A}=\frac{2\left(1-\tan ^{2} \mathrm{~A}\right)}{2 \tan \mathrm{A}}=\tan \mathrm{A}+\frac{1-\tan ^{2} \mathrm{~A}}{\tan \mathrm{A}}=\frac{\tan ^{2} \mathrm{~A}+1-\tan ^{2} \mathrm{~A}}{\tan \mathrm{A}}=\frac{1}{\tan \mathrm{A}}=\cot \mathrm{A}
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