Math, asked by tanavdoda67, 6 hours ago

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

Answers

Answered by anindyaadhikari13
11

\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)
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