prove that cog7 alpha + cos 3 alpha - cos 5 alpha - cos alpha / sin 7 alpha - sin 3 alpha - sin 5 alpha + sin alpha = cot 2 alpha
Answers
Step-by-step explanation:
To prove:
\mathsf{\dfrac{cos7\alpha+cos3\alpha-cos5\alpha-cos\alpha}{sin7\alpha-sin3\alpha-sin5\alpha+sin\alpha}=cot2\alpha}sin7α−sin3α−sin5α+sinαcos7α+cos3α−cos5α−cosα=cot2α
\underline{\textsf{Solution:}}Solution:
\textsf{Consider,}Consider,
\mathsf{\dfrac{cos7\alpha+cos3\alpha-cos5\alpha-cos\alpha}{sin7\alpha-sin3\alpha-sin5\alpha+sin\alpha}}sin7α−sin3α−sin5α+sinαcos7α+cos3α−cos5α−cosα
\mathsf{=\dfrac{cos7\alpha+cos3\alpha-(cos5\alpha+cos\alpha)}{sin7\alpha-sin3\alpha-(sin5\alpha-sin\alpha)}}=sin7α−sin3α−(sin5α−sinα)cos7α+cos3α−(cos5α+cosα)
\textsf{Using the identities,}Using the identities,
2cos5αcos2α−2cos3αcos2α
\mathsf{=\dfrac{2\,cos2\alpha(cos5\alpha-cos3\alpha)}{2\,sin2\alpha(cos5\alpha-cos3\alpha)}}=2sin2α(cos5α−cos3α)2cos2α(cos5α−cos3α)
\mathsf{=\dfrac{cos2\alpha}{sin2\alpha}}=sin2αcos2α
\mathsf{=cot\,2\alpha}=cot2α
\implies\boxed{\mathsf{\dfrac{cos7\alpha+cos3\alpha-cos5\alpha-cos\alpha}{sin7\alpha-sin3\alpha-sin5\alpha+sin\alpha}=cot\,2\alpha}}⟹sin7α−sin3α−sin5α+sinαcos7α+cos3α−cos5α−cosα=cot2α
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