Question 52

What is the value of $$\frac{[1 - \tan(90 - \theta)]^2}{[\cos^2(90 - \theta)]} - 1$$?

Solution

$$\frac{[1-\tan(90-\theta)]^2}{[\sec^2(90-\theta)]}-1$$

$$=\frac{[1-\cot(\theta)]^2}{[\cosec^2(\theta)]}-1\ .$$

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

$$=-\sin\left(2\theta\ \right)\ .$$

A is correct choice.

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