Question 52

What is the value of $$\frac{1}{\sin^4 (90 - \theta)}+\frac{1}{[\cos^2 (90 - \theta)] - 1}?$$

Solution

$$\frac{1}{\sin^4(90-\theta)}+\frac{1}{[\cos^2(90-\theta)]-1}\ .$$

$$=\frac{1}{\cos^4\theta}-\frac{1}{1-\sin^2\theta\ }\ .$$

$$=\frac{1}{\cos^4\theta}-\frac{1}{\cos^2\theta\ }\ .$$

$$=\frac{1-\cos^2\theta\ }{\cos^4\theta}\ .$$

$$=\frac{\sin^2\theta\ }{\cos^4\theta}\ .$$

$$=\tan^2\theta\ \sec^2\theta\ .$$

A is correct choice.

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