Question

Derive the expression $1+ta{n}^{2}x=se{c}^{2}x$

Answer 1

Use the trigonometric identities:

We know the identity${\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1——-\left(i\right)$

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

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

We know that

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

$\frac{{\cos }^2\left(x\right)}{{\cos }^2\left(x\right)}=1$

So the equation (i) after substituting becomes

${\tan }^2\left(x\right)+1=\frac{1}{{\cos }^2\left(x\right)}——–\left(ii\right)$

Now we know that $\frac{1}{{\cos }^2\left(x\right)}=se{c}^2\left(x\right)$

So on substitution equation (ii) we get

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

Hence derived.