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Question
Prove the following:
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution
L.H.S. = tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A
= `tan"A" + 2tan2"A" + 4tan4"A" + 8/(tan8"A")`
= `tan"A" + 2tan2"A" + 4tan4"A" + (8(1 - tan^2 4"A"))/(2tan4"A") ...[because tan2theta = (2tantheta)/(1 - tan^2theta)]`
= `tan"A" + 2tan2"A" + (8tan^2 4"A" + 8 - 8tan^2 4"A")/(2tan4"A")`
= `tan"A" + 2tan2"A" + 4/(tan4"A")`
= `tan"A" + 2tan2"A" + (4(1 - tan^2 2"A"))/(2tan2"A")`
= `tan"A" + (4tan^2 2"A" + 4 - 4tan^2 2"A")/(2tan2"A")`
= `tan"A" + 2/(tan2"A")`
= `tan"A" + (2(1 - tan^2"A"))/(2tan"A")`
= `(2tan^2"A" + 2 - 2tan^2"A")/(2tan"A")`
= `1/tan"A"`
= cot A
= R.H.S.
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