[tex]\large\underline{\sf{Solution-}}[/tex]
We have to prove that:
[tex]\rm\longmapsto \dfrac{ \sin(x) - 2 \sin^{3} (x) }{2 \cos^{3} (x) - \cos(x) } = \tan(x) [/tex]
Taking LHS, we get:
[tex]\rm = \dfrac{ \sin(x) - 2 \sin^{3} (x) }{2 \cos^{3} (x) - \cos(x) }[/tex]
We can take sin(x) as common from numerator part. We get:
[tex]\rm = \dfrac{ \sin(x) \{1- 2 \sin^{2} (x) \}}{2 \cos^{3} (x) - \cos(x) }[/tex]
Similarly, we can take cos(x) as common from first two terms. We get:
[tex]\rm = \dfrac{ \sin(x) \{1- 2 \sin^{2} (x) \}}{ \cos(x) \{ 2 \cos^{2} (x) -1\}}[/tex]
[tex]\rm = \dfrac{ \sin(x) }{ \cos(x) } \cdot \dfrac{\{1- 2 \sin^{2} (x) \}}{\{ 2 \cos^{2} (x) -1\}}[/tex]
As we know that:
[tex]\rm\longmapsto \dfrac{ \sin(x) }{ \cos(x) } = \tan(x) [/tex]
We get:
[tex]\rm = \tan(x) \cdot \dfrac{\{1- 2 \sin^{2} (x) \}}{\{ 2 \cos^{2} (x) -1\}}[/tex]
Now, we will expand the terms in fraction:
[tex]\rm = \tan(x) \cdot \dfrac{\{1- \sin^{2} (x) - { \sin }^{2}(x) \}}{\{\cos^{2} (x) + \cos^{x}(x) -1\}}[/tex]
As we know that:
[tex]\rm \longmapsto \sin^{2}(x)+cos^{2}(x)=1[/tex]
[tex]\rm \longmapsto\cos^{2}(x)=1-sin^{2}(x)[/tex]
[tex]\rm \longmapsto\cos^{2}(x) - 1=-sin^{2}(x)[/tex]
Now substitute the values in the expression, we get:
[tex]\rm = \tan(x)\cdot \dfrac{ \cos^{2} (x)-{ \sin }^{2}(x)}{\cos^{2} (x) - \sin^{2} (x) }[/tex]
[tex]\rm = \tan(x)\cdot1[/tex]
[tex]\rm = \tan(x)[/tex]
Here, we notice that LHS = RHS
Proved.
