Question

in∆ABC angel B=90°
prove it ,sin^2A+sin^2C=1​

Answer 1

Step-by-step explanation:

we all know that

base^2 + perpendicular^2 = hypotenuse^2

in this triangle

AB^2 + BC^2 = AC^2-------------(1)

Now according to the question

sinA = BC/AC, sinC = AB/AC sin^2A + sin^C => (BC/AC)^2 + (AB/AC)^2

=> BC^2/AC^2 + AB^2/AC^2

=> (BC^2 + AB^2)/AC^2

According to (1) - AB^2 + BC^2 = AC^2-----------------(2)

so put the value of AC^2 in (2)

=> (AB^2 + BC^2)/(AB^2 + BC^2)

=> 1