Question

Answer this question please ​

Answer 1

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

We have to prove that:

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

Now consider the first term in LHS. We have:

$\rm = \sqrt{ \dfrac{1 + \cos(x) }{1 - \cos(x) }}$

Multiplying both numerator and denominator by 1 + cos(x), we get:

$\rm = \sqrt{ \dfrac{ \{1 + \cos(x) \} \{1 + \cos(x) \} }{ \{1 - \cos(x) \} \{1 + \cos(x) \}}}$

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

We know that:

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

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

Therefore, we get:

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

$\rm = \dfrac{1 + \cos(x) }{ \sin(x) }$

Now, the second term in LHS is just the reciprocal of first term.

Therefore, the second term will be:

$\rm = \dfrac{ \sin(x)}{1 + \cos(x) }$

Now consider LHS:

$\rm = \dfrac{1 + \cos(x) }{ \sin(x) } + \dfrac{ \sin(x) }{1 + \cos(x) }$

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

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

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

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

Taking 2 as common, we get:

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

$\rm = \dfrac{2}{ \sin(x) }$

We observe that LHS = RHS. (Hence Proved)

$\textsf{\large{\underline{Learn More}:}}$

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

5. Even Odd Identities.

• sin(-x) = -sin(x).
• cos(-x) = cos(x).
• tan(-x) = -tan(x).

Answer 2

Here is your answer in the attachment.

• Hope it helps you
• thanks
• regards, AKASHITEMHEAVEN