Question

1 - cos tetha/
1 + cos tetha =(coto-cosec)2​

Answer 1

$\underline{\underline{\large{\bf{To\:Prove}}}}$

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

$\underline{\underline{\large{\bf{Proof}}}}$

taking LHS

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

multiplying by (1-cos θ) in numerator and denominator

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

using algebraic identity (a + b) (a - b) = a² - b² in denominator

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

using algebraic identity ( a - b )² = a² + b² - 2 ab in numerator

and using trigonometric identity 1 - cos²A = sin²A

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

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

$\rm{LHS= cosec^2\theta + cot^2\theta - 2 cot \theta cosec \theta }$

using algebraic identity ( a - b )² = a² + b² - 2 ab

$\rm{LHS=(cot\theta-cosec\theta)^2 = RHS }$

Proved .