Question

sin cube theta + cos cube theta by sin theta + cos theta + sin theta* cos theta equals to 1

Answer 1

HELLO DEAR,

we know that:- (x³ + y³) = (x + y)(x² + y² - xy)
and (sin²Ф + cos²Ф) = 1

Now the given questions is:- $\bold{\frac{sin^3\theta + cos^3\theta}{sin\theta + cos\theta} + sin\theta*cos\theta}$

now, using the formula of (x³ + y³) in the place of (sin³Ф + cos³Ф).

$\bold{\frac{(sin\theta + cos\theta)(sin^2\theta + cos^2\theta - sin\theta*cos\theta)}{sin\theta + cos\theta} + sin\theta * cos\theta}$

$\bold{[(sin^2\theta + cos^2\theta) - sin\theta * cos\theta] + sin\theta * cos\theta}$

$\bold{1 - sin\theta * cos\theta + sin\theta * cos\theta}$

$\bold{=1}$

$\boxed{\large{\bold{HENCE, \frac{sin^3\theta + cos^3\theta}{sin\theta + cos\theta} + sin\theta*cos\theta = 1}}}$

I HOPE ITS HELP YOU DEAR,
THANKS