Question

prove that sin^2A+cos^2A=1​ on the base of trigonometry​

Answer 1

Answer:

Step-by-step explanation:

Let there be right triangle ABC, 90° at B.

So, using Pythagoras theorem,

$AB^2 + BC^2 = AC^2$

Dividing both sides by $AC^2,$

$\cfrac{AB^2}{AC^2} + \cfrac{BC^2}{AC^2} = \cfrac{AC^2}{AC^2}$

$\Rightarrow \left( \cfrac{AB}{AC} \right)^2 + \left( \cfrac{BC}{AC} \right)^2 = 1$

Since,

$\cfrac{AB}{AC} = \cos A \text{ and } \cfrac{BC}{AC} = \sin A$

$\therefore \cos^2 A + \sin^2 A = 1$

$\Rightarrow \sin^2 A + \cos^2 A = 1$

Proved.

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