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UPSC NDA Math Model Paper 7

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Question : 9 of 120

Marks: +1, -0

What is

\frac{\cos 7 x-\cos 3 x}{\sin 7 x-2 \sin 5 x+\sin 3 x}

equal to?

tan x
cot x
tan 2x
cot 2x

Solution:

As we know,

\sin C+\sin D=2 \sin \left(\frac{C+D}{2}\right)

\cos \left(\frac{C-D}{2}\right)

\cos C-\cos D=-2 \sin \left(\frac{C+D}{2}\right)

\sin \left(\frac{C-D}{2}\right)

=\frac{\cos 7 x-\cos 3 x}{\sin 7 x-2 \sin 5 x+\sin 3 x}

=\frac{-2 \sin \left(\frac{7 x+3 x}{2}\right) \sin \left(\frac{7 x-3 x}{2}\right)}{2 \sin \left(\frac{7 x+3 x}{2}\right) \cos \left(\frac{7 x-3 x}{2}\right)-2 \sin 5 x}

=\frac{-2 \sin 5 x \cdot \sin 2 x}{2 \sin 5 x \cdot \cos 2 x-2 \sin 5 x}

=\frac{-2 \sin 5 x \cdot \sin 2 x}{-2 \sin 5 x[1-\cos 2 x]}

=\frac{\sin 2 x}{1-1+2 \sin ^{2} x}

As,

\cos 2 x=1-2 \sin ^{2} x

\sin 2 x=2 \sin x \cos x

=\frac{2 \sin x \cos x}{2 \sin ^{2} x}=\cot x

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