Question

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

Need prove!! ​

Answer 1

${\underline{\underline{\bf Given}}}$

$\sf\dfrac{cos\theta+1-sin\theta}{cos\theta-1+sin\theta}$

${\underline{\underline{\bf To\:prove}}}$

$\sf \dfrac{1+sin\theta}{cos\theta}$

${\underline{\underline{\bf Solution}}}$

Note:- I've used (X) instead of Ø

$\implies \sf \dfrac{ \dfrac{ \sin(x) - \cos(x) + 1 }{ \cos(x) } }{ \dfrac{ \sin(x) + \cos(x) - 1}{ \cos(x) } } \\ \\ \implies\sf \frac{ \tan(x) - 1 + \sec(x) }{ \tan(x) - \sec(x) + 1 } \\ \\ \implies\sf \frac{ \tan(x) + \sec(x) -( \sec^{2} (x) - \tan^{2} (x) )}{\tan(x) - \sec(x) + 1} \\ \\ \implies \sf \frac{ \tan(x) + \sec(x) - ( \sec(x) + \tan(x) )( \sec(x) - \tan(x) )}{\tan(x) - \sec(x) + 1} \\ \\ \implies\sf \frac{ \tan(x) + \sec(x)(\cancel{ tan(x) - \sec(x) + 1)} }{\cancel{tan(x) - \sec(x) + 1}} \\ \\ \implies\sf \tan(x) + \sec(x) \\ \\ \implies \sf \frac{ \sin(x) }{ \cos(x) } + \frac{1}{ \cos(x) } = \frac{1 + \sin(x) }{ \cos(x) }$

${\underline{\underline{\bf Identities\:used}}}$

$\underline\pink{\boxed{\bf sec²\theta-tan²\theta=1}}$

$\underline\green{\boxed{\bf a²-b²=(a+b)(a-b)}}$