Question

sin^2 alpha * cos^2 beta - cos^2 alpha * sin^2 beta = sin^2 alpha * sin ^2 beta​

Answer 1

Answer:

$\boxed{To Prove}$

$\sf (sin^2 \alpha cos^2 \beta) - (cos^2\alpha sin^2 \beta) = sin^2 \alpha - sin ^2 \beta \\\\\sf$

Now, by formula

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

We can write

$\sf cos^2 \beta = 1-sin^2 \beta ~~ and, cos^2 \alpha = 1-sin^2 \beta$

Now, A/q

$\sf (sin^2 \alpha cos^2 \beta) - (cos^2 \alpha sin^2 \beta) \\\\\sf = sin^2 \alpha(1-sin^2 \beta) - (1-sin^2 \alpha) (sin^2 \beta) \\\\\sf = sin^2 \alpha - sin^2 \alpha sin^2 \beta - (sin^2 \beta - sin^2 \alpha sin^2 \beta) \\\\\sf = sin^2 \alpha - sin^2 \alpha sin^2 \beta - sin^2 \beta + sin^2 \alpha sin^2 \beta \\\\\sf =sin^2 \alpha - \cancel {sin^2 \alpha sin^2 \beta} - sin^2 \beta + \cancel {sin^2 \alpha sin^2 \beta}\\\\\sf = sin^2 \alpha - sin^2 \beta$