Question

prove that: sin^3thetha+cos^3thetha/sin thetha+ cos thetha =1-sin thetha×cos thetha.​

Answer 1

Answer:

wkt

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

(sin +cos)(sin^2-sincos +cos^2)/(sin+cos)

=1(sin^2+cos^2-sincos)

=1-sincos

sorry that I not wrote theta

Answer 2

Step-by-step explanation:

LHS:-

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

• Use algebraic identity

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

$\qquad\qquad\sf:\longrightarrow \dfrac {(sin\theta+cos\theta)(sin\theta^2-sin\theta \times cos\theta +cos\theta^2)}{(sin\theta+cos\theta)}$

$\qquad\qquad\sf:\longrightarrow \dfrac {\cancel {(sin\theta+cos\theta)}(sin^2\theta-sin\theta.cos\theta+cos^2\theta)}{\cancel {(sin\theta+cos\theta)}}$

$\qquad\qquad\sf:\longrightarrow (sin^2\theta-sin\theta.cos\theta+cos^2\theta)$

$\qquad\qquad\sf:\longrightarrow (sin^2\theta+cos^2\theta-sin\theta.cos\theta)$

$\qquad\qquad\sf:\longrightarrow 1-sin\theta\times cos\theta$

$\therefore \underline{\sf LHS=RHS {}_{(proved)}}$.