Question

in triangle abc angle C is equal to 90 degree then prove that sin square b + cos square b is equal to 1

Answer 1

Here is your que's answer dear.

Answer 2

Answer:

Step-by-step explanation:

To prove: $sin^{2}b+cos^{2}b=1$

Proof: From figure,

$sinb=\frac{P}{H}$ and $cosb=\frac{B}{H}$, where P, B and H are the perpendicular, base and hypotenuse of the triangle ABC.

Now, $sin^{2}b+cos^{2}b=(\frac{P}{H}) ^{2}+(\frac{B}{H})^{2}$

$sin^{2}b+cos^{2}b=\frac{P^{2} }{H^{2}}+\frac{B^{2}}{H^{2}}$

$sin^{2}b+cos^{2}b=\frac{P^{2}+B^{2}}{H^{2} }$

Also, From pythagoras theorem, $(Hypotenuse)^{2}=(Perpendicular)^{2}+(Base)^{2}$,

$sin^{2}b+cos^{2}b=\frac{H^{2}}{H^{2} }$

$sin^{2}b+cos^{2}b=1$