Simplify $\frac {\sec^2\theta - \cos^2\theta}{\tan^2\theta}$

Question

My question is to simplify:$$\frac {\sec^2\theta - \cos^2\theta}{\tan^2\theta}$$

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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.

Answer 1

$$\frac{\sec^2 \theta-\cos^2 \theta}{\tan^2 \theta}$$

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

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 $1-\cos^2\theta = \sin^2 \theta$.

$$= \frac{{(\sin^2 \theta)(1+\cos^2 \theta)}}{\sin^2 \theta} = 1+\cos^2 \theta$$

Answer 2

For $\cos \theta \neq 0$ we have

$$\frac {\cos^2\theta}{\cos^2\theta}\cdot \frac {\frac {1}{\cos^2\theta}- \cos^2\theta} {\tan^2\theta}=\frac{1-\cos^4 \theta}{\sin^2\theta}=\frac{(1-\cos^2 \theta)(1+\cos^2 \theta)}{1-\cos^2 \theta}=1+\cos^2 \theta$$

Answer 3

Here is a detailed solution.
Move on! $$\require{cancel} \begin{align} \frac{\sec^2\theta-\cos^2\theta}{\tan^2\theta}&=\frac{\frac{1}{\cos^2\theta}-\cos^2\theta}{\frac{\sin^2\theta}{\cos^2\theta}}\tag{1}\label{ko-eq1}\\ &=\frac{\frac{1-\cos^4\theta}{\cancel{\cos^2\theta}}}{\frac{\sin^2\theta}{\cancel{\cos^2\theta}}}\tag{2}\label{ko-eq2}\\ &=\frac{1-\cos^4\theta}{\sin^2\theta}\tag{3}\label{ko-eq3}\\ &=\frac{\cancel{\left(1-\cos^2\theta\right)}\left(1+\cos^2\theta\right)}{\cancel{1-\cos^2\theta}}\tag{4}\label{ko-eq4}\\ &=1+\cos^2\theta\tag{5}\label{ko-eq5} \end{align}$$ \eqref{ko-eq1} Expand $\sec^2\theta$ as $\frac{1}{\cos^2\theta}$ and $\tan^2\theta$ as $\frac{\sin^2\theta}{\cos^2\theta}$.
\eqref{ko-eq2} Get a common denominator of $\frac{1}{\cos^2\theta}-\cos^2\theta$ and cancel both $\cos^2\theta$. The expression $\frac{\frac{1-\cos^4\theta}{\cos^2\theta}}{\frac{\sin^2\theta}{\cos^2\theta}}$ is equal to $\frac{1-\cos^4\theta}{\cos^2\theta}\cdot\frac{\cos^2\theta}{\sin^2\theta}$.
\eqref{ko-eq3} After cancelling both $\cos^2\theta$.
\eqref{ko-eq4} Expand $1-\cos^4\theta$ as $\left(1-\cos^2\theta\right)\left(1+\cos^2\theta\right)$, change $\sin^2\theta$ as $1-\cos^2\theta$ and cancel the same terms on numerator and denominator.
\eqref{ko-eq5} Final result.
I hope this helps.