Question

plz solve this don't spam, I will mark brainliast for the good explanation ​

Answer 1

Answer:

1

Step-by-step explanation:

a^3+b^3= (a+b)(a^2-ab+b^2)

=(sin©+cos©)(sin^2©-sin©cos©+cos^2©)/(sin©+cos©) + (sin©cos©)

=1-sin©cos©+sin©cos©

1

Answer 2

Given:-

Simplify :-

$\sf \bigg(\dfrac{sin^3\theta + cos^3\theta}{sin\theta + cos\theta}\bigg) + sin\theta cos\theta$

Answer:-

$\sf \bigg(\dfrac{sin^3\theta + cos^3\theta}{sin\theta + cos\theta}\bigg) + sin\theta cos\theta$

${\green{\bigstar}} \boxed{\sf{a^3+b^3 = (a+b)(a^2+b^2-ab)}}$

$\sf \longrightarrow \scriptsize \bigg\{ \dfrac{(sin\theta + cos\theta)(sin^2\theta + cos^2\theta - sin\theta cos\theta)}{sin\theta + cos\theta}\bigg\} + sin\theta cos\theta$

$\sf \longrightarrow \scriptsize \bigg\{ \dfrac{{\cancel{(sin\theta + cos\theta)}}(sin^2\theta + cos^2\theta - sin\theta cos\theta)}{{\cancel{sin\theta + cos\theta}}}\bigg\} + sin\theta cos\theta$

${\blue{\bigstar}} \boxed{\sf{sin^2\theta + cos^2\theta = 1}}$

$\sf \longrightarrow 1 - sin\theta cos \theta + sin\theta cos\theta$

$\sf \longrightarrow 1 \quad {\cancel{- sin\theta cos \theta}} \quad {\cancel{ + sin\theta cos\theta}}$

$\longrightarrow \underline{\underline{\sf{\green{\;\;1\;\;}}}}$