# How do you verify cos^2x + tan^2x - 1 / cos^2x + sin^2x = 0?

May 28, 2018

#### Explanation:

We know that,

color(blue)((1)1/costheta=sectheta

color(red)((2)sin^2theta+cos^2theta=1

color(red)((3)sec^2theta-tan^2theta=1

We have to verify :

${\cos}^{2} x + {\tan}^{2} x - \textcolor{b l u e}{\frac{1}{\cos} ^ 2 x} + {\sin}^{2} x = 0$

We take left hand side :

LHS=cos^2x+tan^2x-color(blue)(1/cos^2x)+sin^2x...tocolor(blue)(Apply(1)

$L H S = {\cos}^{2} x + {\tan}^{2} x - \textcolor{b l u e}{{\sec}^{2} x} + {\sin}^{2} x$

$L H S = {\cos}^{2} x + {\sin}^{2} x - {\sec}^{2} x + {\tan}^{2} x$

$L H S = \left\{\textcolor{red}{{\cos}^{2} x + {\sin}^{2} x}\right\} - \left\{\textcolor{red}{{\sec}^{2} x - {\tan}^{2} x}\right\}$

LHS=1-1...tocolor(red)(Apply(2) and (3)

$L H S = 0$

$L H S = R H S$

May 28, 2018

color(green)(=> 0 = R H S

#### Explanation:

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

${\cos}^{2} x + {\sin}^{2} x + {\tan}^{2} x - \left(\frac{1}{\cos} ^ 2 x\right)$, rearranging terms

${\cos}^{2} x + {\sin}^{2} = 1$, identity

$\sec x \equiv \frac{1}{\cos} x$, identity

$\therefore {\cancel{{\cos}^{2} x + {\sin}^{2} x}}^{\textcolor{red}{1}} + {\tan}^{2} x - {\sec}^{2} x$

$1 + {\tan}^{2} x = {\sec}^{2} x$, identity

$\implies \cancel{1 + {\tan}^{2} x} - \cancel{{\sec}^{2} x} = 0$

color(green)(=> R H S