Question

sin theta tan theta /1-cos theta = 1+sec theta​

Answer 1

To prove:

$\large{ \sf {\dfrac{sin \theta \: tan \theta}{1 - cos \theta} = 1 + sec \theta}}\\\\\\\sf{Taking \: the \: LHS}\\\\ \\\sf{\dfrac{sin \theta \times \dfrac{sin \theta}{cos \theta}}{1 - cos \theta}}\\\\\\\\= \sf{\dfrac{ \dfrac{sin^{2} \theta }{cos \theta}}{1 - cos \theta}}\\\\\\\\= \sf{\dfrac{sin^{2} \theta}{cos \theta} \times \dfrac{1}{1 - cos \theta}}\\\\\\\\= \sf{\dfrac{1 - cos^{2} \theta}{cos \theta} \times \dfrac{1}{1 - cos \theta}}$

$= \sf{\dfrac{(1 + cos \theta)(1 -cos \theta)}{cos \theta} \times \dfrac{1}{1 - cos \theta}}\\\\\\\\= \sf{\dfrac{1 + cos \theta}{cos \theta}}\\\\\\= \sf{\dfrac{1}{cos \theta}+ \dfrac{cos \theta}{cos \theta}}\\\\\\ = \sf{sec \theta + 1}\\\\= \boxed{\bf{1 + sec \theta}}\\\\\\\rm{HENCE \: PROVED}$

Identities used:

• tanθ = sinθ ÷ cosθ
• sin²θ = 1 - cos²θ
• (a + b)(a - b) = a² - b²
• 1 ÷ cosθ = secθ