1 Answer
Solution :-
We known that,
sinA+sinB=2sin\( ( \dfrac{A+B}{2}).cos( \dfrac{A-B}{2})\)
cosA+cosB=2cos\( ( \dfrac{A+B}{2}).cos( \dfrac{A-B}{2})\)
And L.H.S.=
\( \dfrac{(sin7x+sin5x)+(sin9x+sin3x)}{(cos7x+cos5x)+(cos9x+cos3x)}\)
Using the formula and simplifying,
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= \( \dfrac{[2sin6x.cosx]+[2sin6x.cos3x]}{[2cos6x.cosx]+[2cos6x.cos6x]}\)
= \(\dfrac{2sin6x[cosx+cos3x]}{2cos6x[cosx+cos3x]}\)
=tan6x
Therefore L.H.S = R.H.S
Hence proved.
Answered by Abhisek | 1 year agoRelated Questions
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