Question

Prove that $\frac{(cos^{3} \theta - cos 3\theta)}{cos \theta} + \frac{(sin^{3} \theta + sin 3\theta)}{sin \theta} = 3$

Answer 1

Answer:

3

Step-by-step explanation:

Formula used:

$cos3A=4{cos}^3A-3cosA$

$sin3A=3sinA-4{sin}^3A$

${cos}^2A+{sin}^2A=1$

$\frac{{cos}^3\theta-cos3\theta}{cos\theta}+\frac{{sin}^3\theta+sin3\theta}{sin\theta}$

$=\frac{{cos}^3\theta-(4{cos}^3\theta-3cos\theta)}{cos\theta}+\frac{{sin}^3\theta+(3sin\theta-4{sin}^3\theta)}{sin\theta}$

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

$=-3{cos}^2\theta+3+3-3{sin}^2\theta$

$=6-3({cos}^2\theta+{sin}^2\theta})\\=6-3(1)\\=6-3\\=3$