Question

10. Prove that sin? A+cos? A=1​

Answer 1

Given to prove that :-

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

Proof:-

As we know that from trigonometric ratios

$\sf sinA=\dfrac{opposite\:side}{hypotenuse}$

$\sf cosA=\dfrac{adjacent\:side}{hypotenuse}$

Now substituting the values in sin²A+cos²A

$\sf\bigg(\dfrac{opposite\:side}{hypotenuse}\bigg)^{2} +\bigg(\dfrac{adjacent\:side}{hypotenuse}\bigg)^{2}$

$\sf\dfrac{(opposite\:side)^2}{(hypotenuse)^2}+\dfrac{(adjacent\:side)^2}{(hypotenuse)^2}$

$\sf\dfrac{(opposite\:side)^2+(adjacent\:side)^2}{(hypotenuse)^2}$

As we know from Pythagoras theorem

(opposite side)²+(adjacent side)²= (hypotenuse)²

So,

$\sf=\dfrac{(hypotenuse)^2}{(hypotenuse)^2}$

$\sf=1$

HENCE PROVED!

Know more formulae :-

Trigonometric Identities:-

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigonometric relations:-

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonometric ratios:-

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj