Question

please solve the above question!! ​

Answer 1

Step-by-step explanation:

hope you understand what you want.

Answer 2

$\huge \underline{ \underline \mathfrak \pink{answer - }}$

To prove =

$\huge \frac{ { \sin }^{3} \theta + { \cos }^{3} \theta}{ \sin \theta + \cos \theta } + \sin \theta \cos \theta = 1$

Proof -

LHS =

$\frac{ { \sin }^{3} \theta + { \cos }^{3} }{ \sin\theta + \cos \theta } + \sin \theta \cos\theta$

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

[(a³+b³) = (a+B) (a²+b²-ab)]

$= (1 - \sin \theta \cos \theta) + \sin \theta \cos\theta = 1 (\therefore { \sin }^{2} \theta + { \cos }^{2} \theta = 1)$

=RHS

$\therefore \: lhs = rhs$