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Simplify :$\large\frac{\sin 7x-\sin 3x-\sin 5x+\sin x}{\cos 7x+\cos 3x-\cos 5x-\cos x}$

$(a)\;\tan 2x\qquad(b)\;\tan x\qquad(c)\;\cot x\qquad(d)\;1$

1 Answer

Numerator =$(\sin 7x+\sin x)-(\sin 5x+\sin 3x)$
$\qquad\qquad=2\sin 4x.\cos 3x-2\sin 4x.\cos x$
Using the formula
$\sin C-\sin D=2\cos\large\frac{C+D}{2}$$.\sin\large\frac{C-D}{2}$
$\Rightarrow (\sin 7x+\sin x)-(\sin 5x+\sin 3x)$
$\Rightarrow \big[(2\sin\large\frac{8x}{2}$$\cos\large\frac{6x}{2})$$-(2\sin\large\frac{8x}{2}.$$\cos\large\frac{2x}{2})\big]$
$\Rightarrow 2\sin 4x.\cos 3x-2\sin 4x.\cos x$
$\Rightarrow 2\sin 4x[\cos 3x-\cos x]$
Denominator :-
$\cos 7x+\cos 3x-\cos 5x-\cos x$
$\Rightarrow 2\sin 4x\sin x-2\sin 4x\sin 3x$
$\Rightarrow 2\sin 4x(\sin x-\sin 3x)$
Now we have
$\Rightarrow \large\frac{2\sin 4x[\cos 3x-\cos x]}{2\sin 4x[\sin x-\sin 3x]}$
$\Rightarrow \large\frac{\cos 3x-\cos x}{\sin x-\sin 3x}$
$\Rightarrow \large\frac{\cos x-\cos 3x}{\sin 3x-\sin x}$
$\Rightarrow \large\frac{2\sin 2x\sin x}{2\cos 2x\sin x}$
$\Rightarrow \large\frac{\sin 2x}{\cos 2x}$
$\Rightarrow \tan 2x$
Hence (a) is the correct answer.
answered Oct 16, 2013 by sreemathi.v

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