Question

(tan5x+tan3x)/(sin5x-cos3x)=4cos2x cos4x

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Answer 1

Correct Statement:-

$\tt :  ⟼ \: \dfrac{tan5x + tan3x}{tan5x - tan3x} = 4cos2x \: cos4x$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

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$\ \boxed {\red{\tt :  ⟼sinxcosy + sinycosx = sin(x + y)}}$

$\ \boxed {\blue{\tt :  ⟼ \: sinxcosy - sinycosx = sin(x - y)}}$

$\boxed {\green{\tt :  ⟼sin2x \: = 2 \: sinx \: cosx}}$

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$\large\underline\purple{\bold{Solution :- }}$

$\tt :  ⟼ \: \dfrac{tan5x + tan3x}{tan5x - tan3x}$

$\tt :  ⟼ \: \dfrac{\dfrac{sin5x}{cos5x} + \dfrac{sin3x}{cos3x} }{\dfrac{sin5x}{cos5x} - \dfrac{sin3x}{cos3x} }$

$\tt :  ⟼ \: \dfrac{\dfrac{sin5xcos3x + sin3xcos5x}{cos5xcos3x} }{\dfrac{sin5xcos3x - sin3xcos5x}{cos5xcos3x} }$

$\tt :  ⟼ \: \dfrac{sin(5x + 3x)}{sin(5x - 3x)}$

$\tt :  ⟼ \: \dfrac{sin8x}{sin2x}$

$\tt :  ⟼ \: \dfrac{2sin4xcos4x}{sin2x}$

$\tt :  ⟼ \: \dfrac{2(2sin2xcos2x)cos4x}{sin2x}$

$\tt :  ⟼ \: 4cos2xcos4x$