Question

Sin theta/1- cos theta +tan theta /1+cos theta =cot theta +sec theta ×cosec theta

Answer 1

$\underline{\underline{\bold{Question:}}}\\\\\\\bold{Prove\:that:}\\\\\tt{\left(\dfrac{sin\theta}{1-cos\theta}\right)+\left(\dfrac{tan\theta}{1+cos\theta}\right)=cot\theta+sec\theta\:cosec\theta.}\\\\\\\\\tt{Solution:}\\\\\\\underline{\textbf{Solving\:L.H.S}}\\\\\\\tt{=\left(\dfrac{sin\theta}{1-cos\theta}\right)+\left(\dfrac{tan\theta}{1+cos\theta}\right)}\\\\\\\boxed{\tt{a^2-b^2=(a+b)(a-b)}}\\\\\\\tt{=\dfrac{sin\theta(1+cos\theta)+tan\theta(1-cos\theta)}{1-cos^2\theta}}$

$\tt{=\dfrac{sin\theta+cos\theta\:sin\theta+tan\theta-tan\theta\:cos\theta}{1-cos^2\theta}}\\\\\\\tt{=\dfrac{sin\theta+cos\theta\:sin\theta+\dfrac{sin\theta}{cos\theta}-\dfrac{sin\theta}{cos\theta}{\times}cos\theta}{1-cos^2\theta}}\\\\\\\tt{=\dfrac{sin\theta+cos\theta\:sin\theta+\dfrac{sin\theta}{cos\theta}-sin\theta}{1-cos^2\theta}}\\\\\\\tt{=\dfrac{cos\theta\:sin\theta+\dfrac{sin\theta}{cos\theta}}{1-cos^2\theta}}$

$\tt{=\dfrac{sin\theta\left(cos\theta+\dfrac{1}{cos\theta}\right)}{sin^2\theta}}\\\\\\\tt{=\dfrac{\dfrac{cos^2\theta+1}{cos\theta}}{sin\theta}}\\\\\\\tt{=\dfrac{cos^2\theta+1}{sin\theta\:cos\theta}}\\\\\\\tt{=\dfrac{cos^2\theta}{sin\theta\:cos\theta}+\dfrac{1}{sin\theta\:cos\theta}}\\\\\\\tt{=cot\theta+sec\theta\:cosec\theta=R.H.S}\\\\\\\mathfrak{Hence\:Proved.}$