Question

sin³theta+sin3 theta/cos³ theta-cos3 theta​

Answer 1

➼ Answer ➼

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

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

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

$=>\frac{-3}{-3}* [\frac{sin\theta}{cos\theta}]*[\frac{sin^{2}\theta-1}{cos^{2}\theta-1} ]$

$=>tan\theta*[\frac{-cos^{2}\theta}{-sin^{2}\theta} ]$

$=>tan\theta*[\frac{cos^{2}\theta}{sin^{2}\theta} ]$

$=>tan\theta*[\frac{1}{tan^{2}\theta} ]$

$=>\frac{tan\theta}{tan^{2}\theta}$

$=>\frac{1}{tan\theta}$

$=>cot\theta$

Hence,

The value of $\frac{sin^{3}\theta+sin3\theta}{cos^{3}\theta-cos3\theta}$ equals to $cot\theta$.

Formulas for thrice of Angles :-

• $sin3\theta = 3sin\theta -4sin^{3}\theta$
• $cos3\theta=4cos^{3}\theta-3cos\theta$
• $tan3\theta=[\frac{3tan\theta-tan^{3}\theta}{1-3tan^{2}\theta} ]$

☆ Hope it Helps ☆

$@NightRead$