Question

$<marquee>SOLVE THIS QUESTION PROPERLY<marquee>$
spammer keep distance.
i will mark u as brainlist
​

Answer 1

Step-by-step explanation:

$\frac{sin \alpha - 2 { \sin }^{3} \alpha }{2 {cos}^{3} \alpha - \cos\alpha } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha) }{cos \alpha (2 {cos}^{2} \alpha - 1) } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha ) }{cos \alpha (2(1 - {sin}^{2} \alpha ) - 1) } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha )}{cos \alpha(2 - 2 {sin}^{2} \alpha - 1) } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha )}{cos \alpha (1 - 2 {sin}^{2} \alpha ) } \\ \\ = \frac{sin \alpha }{cos \alpha } \\ \\ = tan \alpha \: \: \: \: \: \: \: \: (proved)$

____________________________________________

Concept Used:

• cos²A=1-sin²A
• sinA/cosA=tanA

//Hope this will Helped You//

//Please mark it as Brainliest//

Answer 2

$</p><p>\huge \bf \blue {AnsweR}$

$\begin{lgathered}\frac{sin \alpha - 2 { \sin }^{3} \alpha }{2 {cos}^{3} \alpha - \cos\alpha } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha) }{cos \alpha (2 {cos}^{2} \alpha - 1) } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha ) }{cos \alpha (2(1 - {sin}^{2} \alpha ) - 1) } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha )}{cos \alpha(2 - 2 {sin}^{2} \alpha - 1) } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha )}{cos \alpha (1 - 2 {sin}^{2} \alpha ) } \\ \\ = \frac{sin \alpha }{cos \alpha } \\ \\ = tan \alpha \: \: \: \: \: \: \: \: (proved)\end{lgathered}$