Question

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Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to prove that:

$\rm:\longmapsto \dfrac{ \sin(x) - 2 \sin^{3} (x) }{2 \cos^{3} (x) - \cos(x) } = \tan(x)$

Taking Left Hand Side, we get:

$\rm = \dfrac{ \sin(x) - 2 \sin^{3} (x) }{2 \cos^{3} (x) - \cos(x) }$

We can take sin(x) as common from numerator part. We get:

$\rm = \dfrac{ \sin(x) \{1- 2 \sin^{2} (x) \}}{2 \cos^{3} (x) - \cos(x) }$

Similarly, we can take cos(x) as common from first two terms. We get:

$\rm = \dfrac{ \sin(x) \{1- 2 \sin^{2} (x) \}}{ \cos(x) \{ 2 \cos^{2} (x) -1\}}$

$\rm = \dfrac{ \sin(x) }{ \cos(x) } \cdot \dfrac{\{1- 2 \sin^{2} (x) \}}{\{ 2 \cos^{2} (x) -1\}}$

As we know that:

$\rm: \longmapsto \dfrac{ \sin(x) }{ \cos(x) } = \tan(x)$

We get:

$\rm = \tan(x) \cdot \dfrac{\{1- 2 \sin^{2} (x) \}}{\{ 2 \cos^{2} (x) -1\}}$

Now, we will expand the terms in fraction:

$\rm = \tan(x) \cdot \dfrac{\{1- \sin^{2} (x) - { \sin }^{2}(x) \}}{\{\cos^{2} (x) + \cos^{x}(x) -1\}}$

As we know that:

$\rm: \longmapsto \sin^{2}(x)+cos^{2}(x)=1$

$\rm: \longmapsto\cos^{2}(x)=1-sin^{2}(x)$

$\rm: \longmapsto\cos^{2}(x) - 1=-sin^{2}(x)$

Now substitute the values in the expression, we get:

$\rm = \tan(x)\cdot \dfrac{ \cos^{2} (x)-{ \sin }^{2}(x)}{\cos^{2} (x) - \sin^{2} (x) }$

$\rm = \tan(x)\cdot1$

$\rm = \tan(x)$

We observe that Left Hand Side = Right Hand Side.

Hence Proved..!!

$\textsf{\large{\underline{More To Know}:}}$

1. Relationship between sides.

• sin(x) = Height/Hypotenuse.
• cos(x) = Base/Hypotenuse.
• tan(x) = Height/Base.
• cot(x) = Base/Height.
• sec(x) = Hypotenuse/Base.
• cosec(x) = Hypotenuse/Height.

2. Square formulae.

• sin²x + cos²x = 1.
• cosec²x - cot²x = 1.
• sec²x - tan²x = 1

3. Reciprocal Relationship.

• sin(x) = 1/cosec(x).
• cos(x) = 1/sec(x).
• tan(x) = 1/cot(x).

4. Cofunction identities.

• sin(90° - x) = cos(x) and vice versa.
• cosec(90° - x) = sec(x) and vice versa.
• tan(90° - x) = cot(x) and vice versa.