Question

If bcos θ = a, then prove that cosec θ + cot θ = √(b+a)/(b-a).

Answer 1

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

Given That:

$\rm \longrightarrow b \cos(x) = a$

$\rm \longrightarrow \cos(x) = \dfrac{a}{b}$

Now, consider LHS:

$\rm = \csc(x) + \cot(x)$

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

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

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

Now, substitute the value of cos(x), we get:

$\rm = \dfrac{ 1 + \dfrac{a}{b} }{ \sqrt{1 - \dfrac{ {a}^{2} }{ {b}^{2} } }}$

$\rm = \dfrac{\dfrac{a + b}{b} }{ \sqrt{\dfrac{ {b}^{2} - {a}^{2} }{ {b}^{2} } }}$

$\rm = \dfrac{\dfrac{a + b}{b} }{ \dfrac{\sqrt{ {b}^{2} - {a}^{2}}}{b}}$

$\rm = \dfrac{a + b}{\sqrt{ {b}^{2} - {a}^{2}}}$

$\rm = \sqrt{ \dfrac{ {(a + b)}^{2} }{ {b}^{2} - {a}^{2}}}$

$\rm = \sqrt{ \dfrac{(a + b)(a + b) }{(b + a)(b - a)}}$

$\rm = \sqrt{ \dfrac{a + b}{b - a}}$

$\rm = RHS$

Hence Proved!

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

1. Relationship between sides and T-Ratios.

• 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)