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Mathematics Question on Trigonometric Functions

The value of (cos⁡(π2+x)+sin⁡(π2+x)cos⁡(π2−x)−sin⁡(π2−x))2\left(\frac{\cos\left(\frac{\pi}{2} + x\right) + \sin\left(\frac{\pi}{2}+ x\right)}{\cos\left(\frac{\pi}{2} - x\right) - \sin\left(\frac{\pi}{2} - x\right)}\right)^{2} is

A

1−sin⁡ 2x1+sin⁡ 2x\frac{1 - \sin \, 2x}{1 + \sin \, 2x}

B

1+sin⁡ 2x1−sin⁡ 2x\frac{1 + \sin \, 2x}{1 - \sin \, 2x}

C

1

D

None of these

Answer

1

Explanation

Solution

Consider (cos⁡(π2+x)+sin⁡(π2+x)cos⁡(π2−x)−sin⁡(π2−x))2\left(\frac{\cos\left(\frac{\pi}{2} + x\right) + \sin\left(\frac{\pi}{2}+ x\right)}{\cos\left(\frac{\pi}{2} - x\right) - \sin\left(\frac{\pi}{2} - x\right)}\right)^{2} =(−sin⁡x+cos⁡xsin⁡x−cos⁡x)2= \left(\frac{-\sin x + \cos x}{\sin x - \cos x}\right)^{2} = \left\\{ - \left( \frac{\sin x - \cos x}{\sin x - \cos x}\right)\right\\}^{2} =(−1)2=1= \left(-1\right)^{2} = 1