Question

Prove that 1/cosec theta - cot theta = cosec theta + cot theta

Answer 1

Question:

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

Solution:

$\longmapsto \ \sf LHS = \dfrac{1}{cosec\theta - cot\theta}$

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

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

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

Hence Proved.

Answer 2

ExplanaTion:

Refer to the attachment.

Some formulas related trigonometry :

• Cosec²A - Cot²A = 1

• Sec²A - tan²A = 1

• sin²A + cos²A = 1

• 1/cosecA = sinA

• SinA = 1/CosecA

• CotA = 1/tanA

• tanA = 1/CotA

• cosA = 1/SecA

• SecA = 1/CosA

• SinA = cos(90-A)

• CosA = Sin(90-A)

• SecA = cosec(90-A)

• CosecA = sec(90-A)

• cotA = tan(90-A)

• tanA = cot(90-A)