Question

$prove that\\\\sin^{2} ∅+cos^{2}∅= 1$

Answer 1

Answer:

✅Here's your answer

As

Hypotenuse square = Perpendicular square + Base square

Let hypotenuse be AC, base be BC and perpendicular be AB

So By Pythagoras theorem ,

${ac}^{2} = {ab}^{2} + {bc}^{2}$

${ac}^{2} = {ab}^{2} + {bc}^{2} \\ \\ dividing \: the \: equation \: by \: ac \\ \\( \frac{ac}{ac} ) ^{2} = \frac{ab}{ac} ^{2} + \frac{bc}{ac} ^{2} \\ \\ 1 = \sin( {a}^{2} ) + \cos( {a}^{2} )$

Hope it helps ✌✌....

Answer 2

we know that

Sina = perpendicular /hypotenuse

cosa = base /hypotenuse

suppose a right angle triangle whose hypotenuse is h , perpendicular is p and base is b

Sina = P/h

cosa = b/h

using pythagores theorem

h^2 = b^2 + p^2

so

Now take LHS

Sin^2 a + cos^2a

p^2/h^2 + b^2/h^2

(p^2 +b^2)/h^2

h^2/h^2

1

so

LHS = RHS

hence proved