Question

Simplify $\left(se{c}^{2}x\right)-\left({\tan }^{2}x\right)=1$using sine and cosine

Answer 1

Step 1: Use the relation between sine and cosine

We know the identity

${\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1----\left(i\right)$

Dividing throughout the equation by ${\cos }^2\left(x\right)$, We get

$\frac{{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}+\frac{{\cos }^2\left(x\right)}{{\cos }^2\left(x\right)}=\frac{1}{{\cos }^2\left(x\right)}........\left(ii\right)$

Step 2: Use the identity $\tan \theta =\frac{\sin \theta }{\cos \theta }$

We know that $\tan \theta =\frac{\sin \theta }{\cos \theta }$

$\therefore \frac{{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}={\tan }^2\left(x\right)$ and $\frac{{\cos }^2\left(x\right)}{{\cos }^2\left(x\right)}=1$

So the equation $\left(ii\right)$ after substituting becomes

${\tan }^2\left(x\right)+1=\frac{1}{{\cos }^2\left(x\right)}.......\left(iii\right)$

Now we know that $\frac{1}{\cos \left(x\right)}=sec\left(x\right)$

So on substitution in equation$\left(iii\right)$becomes

${\tan }^2\left(x\right)+1=se{c}^2\left(x\right)$

On rearranging the terms we get

$\left(se{c}^{2}x\right)-\left({\tan }^{2}x\right)=1$

Hence Proved.