Question

Find the value of tan A + 2 tan2A + 4 tan4A + 8 cot 8A

• A. tanA
• B. cotA
• C. secA
• D. cosecA

Answer 1

The correct option is B

cotA

Tan A + 2 tan2A + 4 tan4A + 8 cot 8A

We observe that in each term,angle is getting doubled.

8A can be written as 2 $\times$ 4A

tan A + 2 tan2A + 4 tan4A + 8 cot 2 $\times$ 4A

= tan A + 2 tan2A + 4 tan4A + 8 $\times$ $\frac{(1-ta{n}^{2}4A)}{2tan4A}$

= tanA + 2tan2A + $\frac{8.tan4A.tan4A+8-8ta{n}^{2}4A}{2tan4A}$

= tanA + 2tan2A + $\frac{8.ta{n}^{2}4A+8-8ta{n}^{2}4A}{2tan4A}$

= tanA + 2tan2A + $\frac{8}{2tan4A}$

= tanA + 2tan2A + $\frac{4(1-ta{n}^{2}2A)}{2tan2A}$

= tanA + $\frac{4}{2tan2A}$

= tanA + $\frac{4}{tan2A}$ = tanA + $\frac{2(1-ta{n}^{2}A)}{2tanA}$

= $\frac{2ta{n}^{2}A+2-2ta{n}^{2}A}{2tanA}=\frac{2}{2tanA}=cotA$