Question

cos 7x + cos 5x by sin 7x -sin5x = cot x​

Answer 1

Given:

$\sf \Rsh \dfrac{\cos 7x+\cos 5x}{ \sin 7x - \sin 5x } = \cot x$

To Prove:

$\sf \Rsh L.H.S = R.H.S$

Proof:

L.H.S

$\sf \to \dfrac{\cos 7x+\cos 5x}{ \sin 7x - \sin 5x} \: solve \: numerto \: and \: denominator \: seperately$

Take numerator

$\sf \to \cos 7x+\cos 5x \qquad[ \cos x + \cos y = 2 \cos \frac{x + y}{2}\cos \frac{x - y}{2} ]$

Putting x = 7x and y = 5x

$\sf :\implies 2 \cos \frac{x + y}{2}\cos \frac{x - y}{2}$

$\sf :\implies 2 \cos( \frac{7x + 5x}{2})\cos (\frac{7x - 5x}{2})$

$\sf :\implies 2 \cos( \frac{12x}{2})\cos (\frac{2x}{2})$

$\sf :\implies 2 \cos(6x)\cos (x)$

$\sf :\implies 2 \cos 6x\cos x.....(1)$

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$\sf Now, it's \: time \: to \: take \: denomiator$

$\sf \to \sin 7x - \sin 5x$

$\sf [ \sin x - \sin y = 2 \cos \frac{x + y}{2}\sin \frac{x - y}{2} ]$

Putting x = 7x and y = 5x

$\sf :\implies2 \cos \frac{x + y}{2}\sin \frac{x - y}{2}$

$\sf :\implies2 \cos( \frac{7x + 5x}{2})\sin (\frac{7x - 5x}{2} )$

$\sf :\implies 2 \cos( \frac{12x}{2})\sin (\frac{2x}{2})$

$\sf :\implies 2 \cos(6x)\sin (x)$

$\sf :\implies 2 \cos 6x\sin x.....(2)$

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Now, divide equation 1) and 2) we get

$\sf \to \dfrac{\cos 7x+\cos 5x}{ \sin 7x - \sin 5x}$

$\sf \to \dfrac{2 \cos 6x\cos x}{2 \cos 6x\sin x}$

$\sf \to \dfrac{ \cancel{2\cos 6x}\cos x}{\cancel{2 \cos 6x}\sin x}$

$\sf \to \dfrac{\cos x}{sin x}$

$\sf \to \cot x = R.H.S$

Hence, Proved

$\sf \Longrightarrow \dfrac{\cos 7x+\cos 5x}{ \sin 7x - \sin 5x } = \cot x$