Question

Prove the equation  ${\sin ^2}\theta  + {\cos ^2}\theta  = 1$.

Answer 1

 We will use the figure of a right-angled triangle to help us solve the question. We will also keep in mind that $\dfrac{P}{H} = \sin \theta $, $\dfrac{B}{H} = \cos \theta $. Where, $P$ stands for perpendicular, $H$ stands for hypotenuse and $B$ stands for base.

Complete step-by-step answer:

Let ABC be a right angled triangle, with $AB$ as perpendicular, $BC$ as Base and $AC$ as Hypotenuse

Let angle $ACB$ be $\theta $, as shown in the below figure.

We will apply the theorem of Pythagoras theorem i.e. $\text{Perpendicular}^2 + \text{base}^2 = \text{hypotenuse}^2$, 

$ \Rightarrow A{B^2} + B{C^2} = A{C^2}$

Diving the above equation by $A{C^2}$, we will have:

$ \Rightarrow \dfrac{{A{B^2} + B{C^2}}}{{A{C^2}}} = \dfrac{{A{C^2}}}{{A{C^2}}} $

$ \Rightarrow \dfrac{{A{B^2}}}{{A{C^2}}} + \dfrac{{B{C^2}}}{{A{C^2}}} = 1 $

As we know that

$ \Rightarrow \dfrac{P}{H} = \sin \theta  $

$  \Rightarrow \dfrac{B}{H} = \cos \theta $

Now, in the equation $\dfrac{{A{B^2}}}{{A{C^2}}} + \dfrac{{B{C^2}}}{{A{C^2}}} = 1$,

$   \Rightarrow \dfrac{{AB}}{{AC}} = \sin \theta$

$  \Rightarrow \dfrac{{BC}}{{AC}} = \cos \theta $

Thus, we get to the conclusion that

$ \Rightarrow {\sin ^2}\theta  + {\cos ^2}\theta  = 1$

Hence proved.

Note: Pythagoras theorem i.e. $\text{Perpendicular}^2 + \text{base}^2 = \text{hypotenuse}^2$, used in this question, is the most basic theorem in mathematics and to be used in most of the questions with right angled triangles. Do not forget that $\dfrac{P}{H} = \sin \theta ,\dfrac{B}{H} = \cos \theta ,\dfrac{P}{B} = \tan \theta $.