# How would I establish identity $\,\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\,$?

Establish the identity: $$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$

The first step I got was: $$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$

Then it tells me to rewrite the factor $$\cos\theta + \tan\theta$$ in the numerator using reciprocal identities.

How would I do that? Here is what the homework problem look like: enter image description here

• Do you know the reciprocal identities of cos and tan? – Eleven-Eleven Mar 5 '16 at 2:20

Hint: $$\cot\theta + \sec\theta = \frac{1}{\tan\theta} + \frac{1}{\cos\theta} = \frac{\cos\theta + \tan\theta}{\tan\theta\cos\theta}$$

On the one hand

\begin{align} \dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} &= \dfrac{\dfrac{\cos\theta}{\sin\theta} + \dfrac{1}{\cos\theta}}{\cos\theta + \dfrac{\sin\theta}{\cos\theta}} = \dfrac{\cos^2\theta+\sin\theta}{\sin\theta\cos\theta} \div\dfrac{\cos^2\theta+\sin\theta}{\cos\theta} \\ &= \dfrac{\cos^2\theta+\sin\theta}{\sin\theta\cos\theta} \cdot \dfrac{\cos\theta}{\cos^2\theta+\sin\theta}\\ &= \dfrac{1}{\sin\theta}. \end{align}

On the other,

\begin{align} \sec\theta\cot\theta &= \dfrac{1}{\cos\theta}\cdot\dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta} . \end{align}

Thus we have shown that

$$\bbox[1ex, border:2pt solid #e10000]{\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta}$$