Question:
prove that tanA + 2tan2A + 4tan4A + 8cos8A = cot A .
Answer:
We have to prove that
tan A + 2tan 2A + 4tan 4A + 8cot 8A = cot A
=> cot A - tan A - 2tan 2A - 4tan 4A - 8cot 8A = 0
Now, cot A - tan A = 1/tan A - tan A
= (1- tan2 A)/tan A
= 1/{tan A/(1- tan2 A)}
= 2/{2tan A/(1- tan2 A)} {Multiply and divide by 2}
= 2/tan 2A {since tan 2A = 2tan A/(1- tan2 A)}
= 2cot 2A
=> cot A - tan A = 2cot 2A .....................1
Now,
cot A - tan A - 2tan 2A - 4tan 4A - 8cot 8A
= 2cot 2A - 2tan 2A - 4tan 4A - 8cot 8A
= 2(cot 2A - tan 2A) - 4tan 4A - 8cot 8A
= 2{2cot 2(2A)} - 4tan 4A - 8cot 8A ...........From equation 1
= 4cot 4A - 4tan 4A - 8cot 8A
= 4(cot 4A - tan 4A) - 8cot 8A
= 4{2cot 2(4A)} - 8cot 8A ...........From equation 1
= 8cot 8A - 8cot 8A
= 0
So, tan A + 2tan 2A + 4tan 4A + 8cot 8A = cot A
=> cot A - tan A - 2tan 2A - 4tan 4A - 8cot 8A = 0
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