My question is to simplify:$$\frac {\sec^2\theta – \cos^2\theta}{\tan^2\theta}$$
My workings:
$$\frac {\sec^2\theta – \cos^2\theta}{\tan^2\theta}=\frac {\frac {1}{\cos^2\theta}- \cos^2\theta} {\tan^2\theta}$$
Then I am unable to continue, can someone please show me some working outs? Thank you.
Best Answer
$$\frac{\sec^2 \theta-\cos^2 \theta}{\tan^2 \theta}$$
$$=\frac{\frac{1}{\cos^2 \theta}-\cos^2 \theta}{\tan^2 \theta}$$
From here, use the common denominator $\cos^2 \theta$.
$$=\frac{\frac{1}{\cos^2 \theta}-\frac{\cos^4 \theta}{cos^2 \theta}}{\tan^2 \theta}$$
$$=\frac{\frac{1-\cos^4 \theta}{\cos^2 \theta}}{\tan^2 \theta} = \frac{\frac{(1-\cos^2 \theta)(1+\cos^2 \theta)}{\cos^2 \theta}}{\tan^2 \theta} = \frac{\frac{(1-\cos^2 \theta)(1+\cos^2 \theta)}{\cos^2 \theta}}{\frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{{(1-\cos^2 \theta)(1+\cos^2 \theta)}}{\sin^2 \theta}$$
Recall $1-\cos^2\theta = \sin^2 \theta$.
$$= \frac{{(\sin^2 \theta)(1+\cos^2 \theta)}}{\sin^2 \theta} = 1+\cos^2 \theta$$