sin theta minus 2 Sin cube theta bai 2 cos cube theta minus cos theta barabar 10 theta
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for ease
let theta= A
L.H.S
= sin A - 2sin³ A / 2cos³A - cos A
= sin A(1 - 2sin²A) / cos A(2cos²A - 1)
= 1- 2sin² A= cos 2A and, 2cos² -1= cos 2A
= sin A (cos 2A) / cos A (cos 2A) =
= sin A / cos A = 10
= tan A = R.S.H
let theta= A
L.H.S
= sin A - 2sin³ A / 2cos³A - cos A
= sin A(1 - 2sin²A) / cos A(2cos²A - 1)
= 1- 2sin² A= cos 2A and, 2cos² -1= cos 2A
= sin A (cos 2A) / cos A (cos 2A) =
= sin A / cos A = 10
= tan A = R.S.H
now it is done
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