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HCU MCA 2013 Paper

Question : 75 of 75

Marks: +1, -0

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

is equal to

Solution:

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

= 1 + \frac{1+\sin x}{\cos x} - \left( \frac{\cos^2 x}{1+\sin x} + \frac{\cos x}{1-\sin x} \right)

= 1 + \frac{(1+\sin x)}{\cos x} - \frac{\cos x}{(1-\sin^2 x)} \left[ \cos x (1-\sin x) + (1+\sin x) \right]

= 1 + \frac{(1+\sin x)}{\cos x} - \frac{\cos x}{\cos^2 x} \{ \cos x - \sin x . \cos x + 1 + \sin x \}

= 1 + \frac{1}{\cos x} \{ 1 + \sin x - \cos x + \sin x.\cos x - 1 - \sin x \}

= 1 + \frac{1}{\cos x} \{ \sin x . \cos x - \cos x \}

= 1 + \sin x - 1

= \sin x

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