Question

Prove that: 

1.  tanA + cotA = 2cosec2A

2.  cotA − tanA = 2cot2A

Deduce that tanA + 2tan 2A + 4tan 4A + 8 cot8A = cot A and more generally

tan A + 2 tan2A + 2² tan 2² A + … + 2^{n − 1} tan^{2n − 1} A + 2ⁿ cot2ⁿ A = cot A

Answer 1

tanA + cotA = tanA + 1/tanA = 1 + tan²A/tanA

Therefore,

tanA = cotA − 2cot2A   .....(1) 

tan2A = cot2A − 2cot4A [changing A to 2A in Eq. ..(1)]

tan4A = cot4A − 2cot8A [similar change]

Multiplying Eqs. (1) –(3) by 1, 2, 2² and adding, we get

tanA + 2tan2A + 2² tan4A = cotA − 8cot 8A

Hence,

tanA + 2tan²A + 2² tan 2²A + 2³ cot2³A = cotA

The general result can be obtained by repeating the above sequence of steps n times.