Question

cos cube theta + sin cube theta upon cos theta + sin theta + cos cube theta minus sin cube theta upon cos theta minus sin theta equals to 2 prove it

Answer 1

Answer:

2

Step-by-step explanation:

$Concept:\\a^3+b^3=(a+b)(a^2-ab+b^2)\\a^3-b^3=(a-b)(a^2+ab+b^2)$

$\frac{cos^3\theta+sin^3\theta}{cos\theta+sin\theta}+\frac{cos^3\theta-sin^3\theta}{cos\theta-sin\theta}\\\\=\frac{(cos\theta+sin\theta)(cos^2\theta-cos\theta.sin\theta+sin^2\theta)}{cos\theta+sin\theta}+\frac{(cos\theta-sin\theta)(cos^2\theta+cos\theta.sin\theta+sin^2\theta)}{cos\theta-sin\theta}\\\\=(cos^2\theta-cos\theta.sin\theta+sin^2\theta)+(cos^2\theta+cos\theta.sin\theta+sin^2\theta)\\\\=(cos^2\theta+sin^2\theta)+(cos^2\theta+sin^2\theta)\\\\=1+1= 2$

Answer 2

Step-by-step explanation:

• first use the identity of a cube plus b cube then normally solve it at last use first identity of trigonometry that is sin square theta plus cos square theta is equal to 1