Question

Prove that (cot θ + cosec θ - 1 ) / (cot θ - cosec θ + 1) = (1 + cos θ) / sinθ

Answer 1

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Answer 2

Solution :-

$\rm \dfrac{cot \theta + cosec \theta - 1}{cot \theta - cosec \theta + 1} = \dfrac{1 + cos \theta}{sin \theta}$

Consider LHS

$\rm = \dfrac{cot \theta + cosec \theta - 1}{cot \theta - cosec \theta + 1}$

Writing cot Φ in terms of sin Φ and cosΦ

$\rm = \dfrac{ \dfrac{cos \theta}{sin \theta} + \dfrac{1}{sin \theta} - 1}{ \dfrac{cos \theta}{sin \theta} - \dfrac{1}{sin \theta} + 1}$

Taking LCM

$\rm = \dfrac{ \dfrac{cos \theta + 1 - sin \theta}{sin \theta} }{ \dfrac{cos \theta - 1 + sin \theta}{sin \theta} }$

$\rm = \dfrac{cos \theta + 1 - sin \theta}{cos \theta - 1 + sin \theta}$

Rearranging the terms

$\rm = \dfrac{cos \theta - sin \theta + 1}{cos \theta + sin \theta - 1}$

Rationalising the denominator

$\rm = \dfrac{cos \theta -sin \theta + 1}{cos \theta + sin \theta - 1} \times \dfrac{cos \theta + sin \theta + 1}{cos \theta + sin \theta + 1}$

$\rm = \dfrac{cos \theta + 1 - sin \theta}{cos \theta + sin \theta - 1} \times \dfrac{cos \theta + 1+ sin \theta}{cos \theta + sin \theta + 1}$

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

$\rm = \dfrac{cos^{2} \theta + 2cos \theta + cos^{2} \theta}{1+ 2cos \theta sin \theta - 1}$

$\rm = \dfrac{2cos^{2} \theta + 2cos \theta }{2cos \theta sin \theta }$

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

$\rm = \dfrac{1 + cos \theta}{ sin \theta }$

= RHS

LHS = RHS

QED

Identities used to prove

• cos² Φ + sin²Φ = 1

• cos² Φ = 1 - sin² Φ

• x² - y² = (x + y)(x - y)

• (x + y)² = x² + y² + 2xy