Question

Brainly stars and moderators. A question for you all:.

sin 5x-sin 3x / cos 5x+cos 3x =tan 4x
Class 11
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Answer 1

Correct Question :

sin 5x + sin 3x / cos 5x + cos 3x = tan 4x

Given :

• sin 5x + sin 3x / cos 5x + cos 3x

To Prove :

• sin 5x + sin 3x / cos 5x + cos 3x =tan 4x

Solution :

Using Identities :

• sin x + sin y = 2sin x+y/2 cos x-y/2
• cos x + cos y = 2cos x+y/2 cos x-y/2

$\longmapsto\tt{\dfrac{2sin\:\dfrac{5x+3x}{2}\:cos\:\dfrac{5x-3x}{2}}{2cos\:\dfrac{5x+3x}{2}\:cos\:\dfrac{5x-3x}{2}}}$

$\longmapsto\tt{\dfrac{2sin\:\dfrac{{\not{8x}}}{{\not{2}}}\:cos\:\dfrac{{\not{2}}x}{{\not{2}}}}{2cos\:\dfrac{{\not{8x}}}{2}\:cos\:\dfrac{{\not{2}}}{{\not{2}}}}}$

$\longmapsto\tt{\dfrac{{\not{2}}\:sin\:4x\:{\cancel{cos\:x}}}{{\not{2}}\:cos\:4x\:{\cancel{cos\:x}}}}$

$\longmapsto\tt\bf{tan\:4x}$

HENCE PROVED!

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Some more Trigonometric Identities :

• cos x + cos y = 2 cos x+y/2 cos x-y/2
• sin x + sin y = 2 sin x+y/2 cos x-y/2
• sin (-x) = -sin x
• cos (-x) = cos x
• cos (x-y) = cos x cos y + sin x sin y
• sin (x-y) = sin x cos y - cos x sin y

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Answer 2

Correct Question

$\:$

• Prove that : $\sf \dfrac{sin\:5x + sin\:3x}{cos\:5x + cos\:3x} = tan\:4x$

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Given

$\:$

• $\sf \dfrac{sin\:5x + sin\:3x}{cos\:5x + cos\:3x}$

$\:$

To Prove

$\:$

• $\sf \dfrac{sin\:5x + sin\:3x}{cos\:5x + cos\:3x} = tan\:4x$

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Proof

$\:$

$\clubsuit$ Identities Used :

$\:$

$\footnotesize\bigstar\:{\pmb{\underline{\boxed{\bf{\red{sin\:A + sin\:B = 2sin\Bigg\{\dfrac{A + B}{2}\Bigg\}cos\:\Bigg\{\dfrac{A - B}{2}\Bigg\}}}}}}}$

$\:$

$\footnotesize\bigstar\:{\pmb{\underline{\boxed{\bf{\purple{cos\:A + cos\:B = 2cos\Bigg\{\dfrac{A + B}{2}\Bigg\}cos\:\Bigg\{\dfrac{A - B}{2}\Bigg\}}}}}}}$

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$\\ \qquad\quad\:\:\:\::\implies\:\pmb{\sf{L.H.S}}$

$\\ \qquad\quad :\implies\:\sf \dfrac{sin\:5x + sin\:3x}{cos\:5x + cos\:3x}$

$\\ :\implies\:\sf \dfrac{2sin\Bigg\{\dfrac{5x + 3x}{2}\Bigg\}cos\:\Bigg\{\dfrac{5x - 3x}{2}\Bigg\}}{2cos\Bigg\{\dfrac{5x + 3x}{2}\Bigg\}cos\:\Bigg\{\dfrac{5x - 3x}{2}\Bigg\}}$

$\\ :\implies\:\sf \dfrac{2sin\Bigg\{\dfrac{\not{8x}}{\not{2}}\Bigg\}cos\:\Bigg\{\dfrac{\not{2x}}{\not{2}}\Bigg\}}{2cos\Bigg\{\dfrac{\not{8x}}{\not{2}}\Bigg\}cos\:\Bigg\{\dfrac{\not{2x}}{\not{2}}\Bigg\}}$

$\\ \qquad\quad:\implies\:\sf \dfrac{\cancel{2}sin\:4x\:\cancel{cos\:x}}{\cancel{2}cos\:4x\:\cancel{cos\:x}}$

$\\ \qquad\quad\:\:\:\::\implies\:\sf \dfrac{sin\:4x}{cos\:4x}$

$\\ :\implies\:\pmb{\blue{\underline{\boxed{\bf{\pink{tan\:4x}}}}}}\qquad\Bigg\lgroup \because\:{\pmb {\frak {\green {\dfrac{sin\:\rm{A}}{cos\:\rm{A}} = tan\:\rm{A}}}}}\Bigg\rgroup$

$\\ \qquad\quad\:\:\:\::\implies\:\pmb{\sf{R.H.S}}$

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$\qquad\quad\:\:\:\therefore\:\pmb{\underline{\textsf{\textbf{Hence,\:Proved!}}}}$

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$\clubsuit$ Know More :

$\:$

$\pmb{\underline{\bm{\bigstar\:Trigonomentary\:Formulas\::}}}$

$\:$

• sin (–A) = –sin A

• tan (–A) = –tan A

• cot (–A) = –cot A

• cosec (–A) = –cosec A

• cos (–A) = cos A

• sec (–A) = sec A

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