Question

Sin cube + cos cube divide by sin + cos

Answer 1

Answer:

$\frac{ \sin {}^{3} + \cos {}^{3} }{ \sin + \cos } = \sin {}^{2} + \cos {}^{2}$

Answer 2

Answer:

1 - sinθcosθ

Or (2 - sin2θ)/2

Or 1 - sin2θ/2  

Step-by-step explanation:

Given,

E = (sin³θ+cos³θ)/(sinθ+cosθ)

Using a³ + b³ = (a+b)(a² - ab +b²)

Here a= sin³θ and b= cos³θ

∴ E = (sinθ + cosθ)(sin²θ - sinθcosθ + cos²θ)/(sinθ + cosθ)

= E = sin²θ - sinθcosθ + cos²θ

∵ sin²θ + cos²θ = 1

= E = 1 - sinθcosθ                        {Solving for class 10th}

This should be the answer in θ.

But we can also reduce it to single trigonometric ratio.

(Multiply and divide expression by 2)

∴ E = (2 - 2sinθcosθ)/2

∵ 2sinθcosθ = sin2θ                            {Solving for class 11th }

= E = (2 - sin2θ)/2 or 1 - sin2θ/2  

∴ Answer = 1 - sinθcosθ or (2 - sin2θ)/2 or 1 - sin2θ/2  

Hope, you got it :-))

Please mark it as brainiest!!