Question 6

Maths Question: 

$\text{Establish that }\frac{{{\tan }^{2}}\theta +{{\cos }^{2}}\theta }{\sec \theta +\sin \theta }=\sec \theta -\sin \theta$

Maths Solution: 

$\begin{align}  & \text{From trig identities} \\ & {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta  \\ & {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 \\ & \text{From the left hand side} \\ & \frac{{{\tan }^{2}}\theta +{{\cos }^{2}}\theta }{\sec \theta +\sin \theta }=\frac{{{\sec }^{2}}\theta -1+{{\cos }^{2}}\theta }{\sec \theta +\sin \theta } \\ & \frac{{{\tan }^{2}}\theta +{{\cos }^{2}}\theta }{\sec \theta +\sin \theta }=\frac{{{\sec }^{2}}\theta +{{\cos }^{2}}\theta -1}{\sec \theta +\sin \theta } \\ & \text{Also from trig identities} \\ & {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta  \\ & -{{\sin }^{2}}\theta ={{\cos }^{2}}\theta -1 \\ & \frac{{{\tan }^{2}}\theta +{{\cos }^{2}}\theta }{\sec \theta +\sin \theta }=\frac{{{\sec }^{2}}\theta -{{\sin }^{2}}\theta }{\sec \theta +\sin \theta } \\ & \frac{{{\tan }^{2}}\theta +{{\cos }^{2}}\theta }{\sec \theta +\sin \theta }=\frac{(\sec \theta -\sin \theta )(\sec \theta +\sin \theta )}{\sec \theta +\sin \theta } \\ & \frac{{{\tan }^{2}}\theta +{{\cos }^{2}}\theta }{\sec \theta +\sin \theta }=\sec \theta -\sin \theta  \\\end{align}$

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