Question

Prove that
(cos A+cos 3A+cos 5A +cos 7A)/(sin A+sin 3A+sin 5A+ sin 7A)=cot 4A​

Answer 1

Step-by-step explanation:proved below in the image

Answer 2

$\large{\underline{\mathrm{\red{Question-}}}}$

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Prove that -

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$\sf{\dfrac{CosA\:+\:Cos3A\:+\:Cos5A\:+\:Cos7A}{SinA\:+\:Sin3A\:+\:Sin5A\:+\:Sin7A}\:=\:Cot4A}$

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$\large{\underline{\mathrm{\red{Solution-}}}}$

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: $\implies$ Taking LHS,

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: $\implies$ $\sf{\dfrac{CosA\:+\:Cos3A\:+\:Cos5A\:+\:Cos7A}{SinA\:+\:Sin3A\:+\:Sin5A\:+\:Sin7A}}$

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: $\implies$ $\sf{\dfrac{(CosA\:+\:Cos7A)\:+\:(Cos3A\:+\:Cos5A)}{(SinA\:+\:Sin7A)\:+\:(Sin3A\:+\:Sin5A)}}$

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: $\implies$ $\sf{\dfrac{2Cos4ACos3A+2Cos4ACosA}{2Sin4ACos3A+2Sin4ACosA}}$

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: $\implies$ $\sf{\dfrac{2Cos4A\:\cancel{(Cos3A+CosA)}}{2Sin4A\:\cancel{(Cos3A+CosA)}}}$

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: $\implies$ $\sf{\dfrac{2Cos4A}{2Sin4A}}$

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°•° $\sf{\dfrac{CosA}{SinA}=CotA}$

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: $\implies$ Cot4A

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: $\implies$ RHS

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Hence proved!