Question

Prove that cos theta /1-sin theta + 1-sin theta/ cos theta = 2 sec theta​

Answer 1

$\sf\huge\red{\underbrace{ Question : }}$

Prove that :

$\sf \cfrac{\cos\:\theta}{1-\sin\:\theta} + \cfrac{1-\sin\:\theta}{\cos\:\theta} = 2\sec\:\theta$

$\sf\huge\orange{\underbrace{ Solution : }}$

$\tt\blue{\underline{ \mapsto \: LHS :}}$

$\sf \implies \cfrac{\cos\:\theta}{1-\sin\:\theta}+\cfrac{1-\sin\:\theta}{\cos\:\theta}$

$\sf \implies \cfrac{\cos^{2}\: \theta + (1-\sin\:\theta)^{2}}{(1-\sin\:\theta)\cos\:\theta}$

• (a - b)² = a² + b² - 2ab

$\sf \implies \cfrac{\cos^{2}\: \theta + (1)^{2} + (\sin\:\theta)^{2} - 2\sin\:\theta}{\cos\:\theta - \sin\: \theta\cos\:\theta}$

$\sf \implies \cfrac{\cos^{2}\:\theta + 1 + \sin^{2}\:\theta - 2\sin\:\theta}{\cos\:\theta - \sin\:\theta\cos\:\theta}$

• sin² θ + cos² θ = 1

$\sf \implies \cfrac{1 + 1 - 2\sin\:\theta}{(1-\sin\:\theta)\cos\:\theta}$

$\sf \implies \cfrac{2- 2\sin\:\theta}{(1-\sin\:\theta)\cos\:\theta}$

$\sf \implies \cfrac{2(1-\sin\:\theta)}{(1-\sin\:\theta)\cos\:\theta}$

$\sf \implies \cfrac{2\cancel{(1-\sin\:\theta)}}{\cancel{(1-\sin\:\theta)}\cos\:\theta}$

$\sf \implies \cfrac{2}{\cos\:\theta}$

$\sf \implies 2\cfrac{1}{\cos\:\theta}$

• 1/cos θ = sec θ

$\sf \implies 2\sec \:\theta \: \: \: \tt\green{=RHS}$

$\blue{\bigstar}\underline{\boxed{\rm{\purple{\therefore Hence,\:it\:was\:proved.}}}}\:\blue{\bigstar}$

$\rm\pink{\underline{ \mapsto More\:information :}}$

$\boxed{\begin{minipage}{7 cm} Trigonometric Identities : \\ \\ \sin^{2}\theta + cos^{2}\theta = 1 \\ \\ 1 + tan^{2}\theta = sec^{2}\theta \\ \\1 + cot^{2}\theta=\text{cosec}^2\, \theta \end{minipage}}$

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