Question

slove out the given attachment

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Answer 1

Given to prove that :-

$\sf\dfrac{(sin7x + sin5x) + (sin9x + sin3x)}{(cos7x + cos5x) + (cos9x + cos3x)} = tan6x$

Concept to know:-

Here the some Trigonometric formulae were used that were :-

$\sf sinC + sinD = 2sin \bigg( \dfrac{C + D}{2} \bigg)cos \bigg( \dfrac{C - D}{2} \bigg)$

$\sf cosC + cosD= 2cos \bigg( \dfrac{C + D}{2} \bigg)cos \bigg( \dfrac{C - D}{2} \bigg)$

By using these formulae we shall simplify the numerator and denominator

Proof :-

By applying these formulae we get ,

$\sf = \dfrac{2sin \bigg(\dfrac{7x + 5x}{2} \bigg)cos \bigg( \dfrac{7x - 5x}{2} \bigg) + 2sin \bigg( \dfrac{9x + 3x}{2} \bigg)cos \bigg( \dfrac{9x - 3x}{2} \bigg) }{2cos \bigg( \dfrac{7x + 5x}{2} \bigg)cos \bigg( \dfrac{7x - 5x}{2} \bigg) + 2cos \bigg( \dfrac{9x + 3x}{2} \bigg)cos \bigg( \dfrac{9x - 3x}{2} \bigg) }$

$\sf = \dfrac{2sin \bigg(\dfrac{12x}{2} \bigg)cos \bigg( \dfrac{2x}{2} \bigg) + 2sin \bigg( \dfrac{12x}{2} \bigg)cos \bigg( \dfrac{6x}{2} \bigg) }{2cos \bigg( \dfrac{12x}{2} \bigg)cos \bigg( \dfrac{2x}{2} \bigg) + 2cos \bigg( \dfrac{12x}{2} \bigg)cos \bigg( \dfrac{6x}{2} \bigg) }$

$\sf = \dfrac{2sin6xcosx +2sin6xcos3x }{2cos6xcosx +2cos6xcos3x }$

Take common 2sin6x in the numerator and 2cos6x in the denominator

$\sf = \dfrac{2sin6x(cosx + cos3x)}{2cos6x(cosx + cos3x)}$

$\sf = \dfrac{2sin6x \not(cos4x)}{2cos6x \not(cos4x)}$

$\sf = \dfrac{2sin6x}{2cos6x}$

$\sf = \dfrac{sin6x}{cos6x}$

From Trigonmetric relations,

sinA/cosA = tanA

$\sf = tan6x$

Hence proved !

Note :-

Kindly swipe the page to right in order to sèe the full answer.

Know more formulae:-

❒Multiple angles :-

$\sf sin2A = 2sinAcosA$

$\sf sin2A = \dfrac{2tanA}{1 + tan {}^{2}A }$

$\sf cos2A = cos {}^{2} A - sin { }^{2} A$

$\sf cos2A= 2cos {}^{2} A - 1$

$\sf cos2A = 1 - 2in {}^{2} A$

$\sf cos2A = \dfrac{1 - tan {}^{2} A}{1 + tan {}^{2}A }$

$\sf tan2A = \dfrac{2tanA}{1 - tan {}^{2}A }$

$\sf cot2A = \dfrac{cot {}^{2} A - 1}{2cotA}$

$\sf sin3A = 3sinA - 4sin {}^{3} A$

$\sf cos3A = 4cos {}^{3} A - 3cosA$

$\sf tan3A = \dfrac{3tanA - tan {}^{3}A }{1 - 3tan {}^{2} A}$

❒Compound angles:-

$\sf sin(A+B)= sinAcosB + sinBcosA$

$\sf sin(A-B) = sinAcosB- sinBcosA$

$\sf cos(A+B) = cosAcosB - sinAsinB$

$\sf cos(A-B) = cosAcosB + sinAsinB$

$\sf tan(A+B) = \dfrac{tanA+tanB}{1-tanAtanB}$

$\sf tan(A-B) = \dfrac{tanA-tanB}{1+tanAtanB}$

$\sf cot(A+B) = \dfrac{cotB cotA -1}{cotB + cotA}$

$\sf cot(A-B) = \dfrac{cotB cotA + 1}{cotB-cotA}$