Question

sec^theeta=1+tan^theeta
​

Answer 1

This is an identity......

Answer 2

To Prove :

$\tt\dagger \: \: \: \: \: 1 + {tan}^{2} \theta= {sec}^{2} \theta.$

Solution :

We have,

• Right Angle Triangle ABC.

By Pythagoras theorem, We can say that.

$\tt \dagger \: \: \: \: \: {H}^{2} = {P}^{2} + {B}^{2}$

Where as,

• Hypotenuse = AC
• Perpendicular = AB
• Base = BC

Let Divided both sides by H², We get.

$\tt\longmapsto \dfrac{{H}^{2}}{{H}^{2} }= \dfrac{{P}^{2}}{{H}^{2}} + \dfrac{{B}^{2}}{ {H}^{2}}$

Note :

$\tt\bullet \: \: sin\theta = \dfrac{Perpendicular}{Hypotenuse}$

$\tt\bullet \: \: cos\theta = \dfrac{Base}{Hypotenuse}$

$\tt\bullet \: \: tan\theta = \dfrac{Perpendicular}{Base}$

So,

$\tt\longmapsto 1 = {sin}^{2} \theta + {cos}^{2} \theta.$

$\tt\longmapsto {sin}^{2} \theta + {cos}^{2} \theta = 1.$

$\rule{100}2$

Now,

$\tt\dagger \: \: \: \: \: {sin}^{2} \theta + {cos}^{2} \theta = 1.$

Divide both sides by Cos theta, We get.

$\tt\longmapsto \dfrac{{sin}^{2} \theta}{{cos}^{2} \theta} + \dfrac{{cos}^{2} \theta }{{cos}^{2} \theta} = \dfrac{1}{{cos}^{2}\theta}$

$\tt\longmapsto {tan}^{2} \theta + 1 = {sec}^{2} \theta.$

$\tt\longmapsto 1 + {tan}^{2} \theta= {sec}^{2} \theta.$

Hance Proved.

$\rule{200}3$

$\boxed{ \begin{minipage}{6 cm} Fundamental Trigonometric Identities \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\ 1 +\tan^2\theta = \sec^2\theta \\ \\ 1 +\cot^2\theta = \text{cosec}^2 \, \theta \end{minipage}}$