Question

If 1 + cosec theta + cot theta / cosec theta - cot theta = ? ​

Answer 1

Question:

\sf Prove \ that: \ \dfrac{1}{cosec\theta - cot\theta} = cosec\theta + cot\thetaProve that:

cosecθ−cotθ

1

=cosecθ+cotθ

Solution:

\begin{gathered}\longmapsto \ \sf LHS = \dfrac{1}{cosec\theta - cot\theta}\\ \\\end{gathered}

⟼ LHS=

cosecθ−cotθ

1

Multiplying the numerator and denominator by the conjugate of (cosecθ - cotθ) i.e, (cosecθ + cotθ) we get,

\begin{gathered}\longmapsto \ \sf LHS = \dfrac{1}{cosec\theta - cot\theta} \times \dfrac{\left(cosec\theta + cot\theta\right)}{\left(cosec\theta + cot\theta\right)}\\ \\ \\ \\\longmapsto \ \sf LHS = \dfrac{cosec\theta + cot\theta}{\left(cosec\theta - cot\theta\right) \left(cosec\theta + cot\theta\right)}\\ \\ \\ \\\sf Using \ the \ identity \ (a + b)(a - b) = a^2 - b^2 we get,\\ \\ \\ \\\longmapsto \ \sf LHS = \dfrac{cosec\theta + cot\theta}{cosec^2\theta - cot^2\theta}\\ \\ \\ \\\end{gathered}

⟼ LHS=

cosecθ−cotθ

1

×

(cosecθ+cotθ)

(cosecθ+cotθ)

⟼ LHS=

(cosecθ−cotθ)(cosecθ+cotθ)

cosecθ+cotθ

Using the identity (a+b)(a−b)=a

2

−b

2

weget,

⟼ LHS=

cosec

2

θ−cot

2

θ

cosecθ+cotθ

\begin{gathered}\\\end{gathered}

\begin{gathered}\sf Using \ the \ identity \ cosec^2\theta - cot^2\theta = 1 \ we \ get,\\ \\ \\ \\\longmapsto \ \sf LHS = \dfrac{cosec\theta + cot\theta}{1}\\ \\ \\ \\\longmapsto \ \sf LHS = cosec\theta + cot\theta\\ \\ \\ \\\longmapsto \sf \ \underline{\underline{LHS = RHS}}\end{gathered}

Using the identity cosec

2

θ−cot

2

θ=1 we get,

⟼ LHS=

1

cosecθ+cotθ

⟼ LHS=cosecθ+cotθ

⟼

LHS=RHS