Question

Prove That,
Sin²A + Cos²B = 1

Answer 1

Question:

Prove That,

Sin²A + Cos²B = 1

Solution:

LHS:

Sin²A + Cos²B

= { $\frac{Perpendicular}{Hypotenuse}$×$\frac{Hypotenuse}{Perpendicular}$ }² + { $\frac{Perpendicular}{Base}$×$\frac{Base}{Perpendicular}$ }²

= {$\frac{BC}{AC}$×$\frac{AC}{BC}$}²+{$\frac{AB}{AC}$×$\frac{AC}{AB}$}²

Answer 2

Answer:

L.H.S.

sin²a cos²b -cos²Asin²B

SIN²A(1-sin²B)-(1-sin²A)sin²B

sin²a -sin²Asin²B -sin²B +sin²Asin²B

sin²a -sin²B

R.H.S. PROVED

Step-by-step explanation:

$Hope \: it \: helps...!!$