Question 54

What is the value of $$\sin(180 - \theta) \sin(90 - \theta) - \left[\frac{\cot(90 - \theta)}{1 + \tan^2 \theta}\right]$$

Solution

$$\sin(180-\theta)\sin(90-\theta)-\left[\frac{\cot(90-\theta)}{1+\tan^2\theta}\right]$$

$$=\sin\theta\ \cos\theta-\left[\frac{\tan\theta}{\sec^2\theta}\right]$$

$$=\sin\theta\ \cos\theta-\left[\frac{\sin\theta}{\cos\theta}.\ \frac{\cos^2\theta\ }{1}\right]$$

$$=\sin\theta\ \cos\theta-\left[\sin\theta\ \cos\theta\ \right]$$

$$=0\ .$$

D is correct choice.

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