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Simplify : $\large\frac{\sin\big(\Large\frac{3\pi}{2}-\theta\big)\cos\big(\Large\frac{\pi}{2}+\theta\big)}{\tan\big(\Large\frac{\pi}{2}+\theta\big)}-\frac{\sin\big(\Large\frac{3\pi}{2}-\theta\big)}{\sec\big(\pi+\theta\big)}$

$(a)\;1\qquad(b)\;0\qquad(c)\;2\qquad(d)\;-1$

1 Answer

The expression can be written as $\large\frac{(-\cos \theta)(-\sin\theta)}{-\cot\theta}-\large\frac{-\cos\theta}{-\sec\theta}$
$\Rightarrow \large\frac{\cos\theta\sin\theta}{-\cot\theta}$$-(-\cos\theta)(-\cos\theta)$
$\Rightarrow \large\frac{\cos\theta\sin\theta}{-\Large\frac{\cos\theta}{\sin\theta}}$$-\cos^2\theta$
$\Rightarrow \large\frac{\cos\theta.\sin\theta.\sin\theta}{-\cos\theta}$$-\cos^2\theta$
$\Rightarrow -\sin^2\theta-\cos^2\theta$
$\Rightarrow -(\sin^2\theta+\cos^2\theta)$
We know that $\sin^2\theta+\cos^2\theta=1$
$\Rightarrow -1$
Hence (d) is the correct answer.

 

answered Oct 16, 2013 by sreemathi.v
edited Jan 6, 2014 by meenakshi.p
 

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