Question

prove it:-
and please don't meaningless
Answer ❎❎​

Answer 1

$\Large{\underline{\tt{\red{\:Solution}}}}$

$\Large{\underline{\tt{\red{\:Important\:Formula}}}}$

★ Sin A + Sin B = 2 * Sin (A+B)/2 * Cos (A- B)/2

★ Cos A + Cos B = 2 * Cos (A+B)/2 * Cos (A-B)/2

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Take L.H.S. ,

[(Sin 7x + Sin 5x) + (Sin 9x + Sin 3x)]/[(Cos 7x + Cos 5x) + (Cos 9x + Cos 3x)]

$\Large{\underline{\tt{\red{\:Using\:Formula}}}}$

= [{2* Sin(7x+5x)/2 * Cos (7x-5x)/2 } + { 2 * Sin (9x+3x)/2 * Cos (9x-3x)/2 }] / [{2 * Cos (7x +5x)/2 * Cos (7x-5x)/2 } + { 2 * Cos(9x+3x)/2 * Cos (9x-3x/2}]

= 2 * [ Sin (12x/2) * Cos (2x/2) + Sin (12x/2) * Cos (6x/2) ] / 2 *[ Cos(12x/2) * Cos (2x/2) + Cos(12x/2) * Cos (6x/2) ]

= [ Sin (6x) * Cos (x) + Sin (6x) * Cos (3x) ] / [ Cos(6x) * Cos (x) + Cos(6x) * Cos (3x) ]

= Sin(6x) * [ Cos x + Cos 3x ] / Cos 6x * [Cos x + Cos 6x ]

= Sin 6x / Cos 6x

We know,

$\small{\tt{\green{\:\:\:\:\:\:\star\tan A\:=\:\dfrac{\sin A}{\cos A}}}}$

So,

= tan 6x .

= R.H.S.

That's proved.

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Answer 2

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