Question

$\sin( \alpha ) - \cos( \alpha ) + 1 \div \sin( \alpha + \cos( \alpha) - 1 = 1 \div \sec( \alpha ) - \tan( \alpha)$
​

Answer 1

Question :-

Prove that ,

$\sf\frac{ \sin \alpha - \cos \alpha + 1}{ \sin \alpha + \cos \alpha - 1} = \frac{1}{ \sec \alpha - \tan \alpha}$

Proof :-

Formula used :-

$\star \: \frac{1}{ \cos \theta} = \sec \theta \\ \\ \star \: \frac{ \sin \theta}{ \cos \theta} = \tan \theta \\ \\ \star \sf \: { \sec}^{2} \theta - { \tan}^{2} \theta = 1$

Explanation :-

We need to prove Left hand side is equal to right hand side ,

Taking left hand side (LHS)

$\rightarrow \: \sf\frac{ \sin \alpha - \cos \alpha + 1}{ \sin \alpha + \cos \alpha - 1} \: \\ \\ \sf \: taking \: \cos \alpha \: common \: in \: both \\ \sf numenator \: and \: denomenator \\ \\ \rightarrow \: \sf\frac{ \tan \alpha - 1 + \sec \alpha}{ \tan \alpha + 1 - \sec \alpha} \: \\ \\ \rightarrow \sf\frac{ \tan \alpha - 1 + \sec \alpha}{ \tan \alpha - \sec \alpha \: + { \sec}^{2} \alpha - { \tan}^{2} \alpha } \: \: \\ \\ \rightarrow \sf\frac{ \tan \alpha - 1 + \sec \alpha}{ (\tan \alpha - \sec \alpha) + ( \sec \alpha + \tan \alpha)( \sec \alpha - \tan \alpha)} \: \: \\ \\ \rightarrow \frac{ \tan \alpha - 1 + \sec \alpha}{( \tan \alpha - \sec \alpha)( - 1 + \tan \alpha + \sec \alpha)} \\ \\ \rightarrow \: \frac{1}{ \tan \alpha - \sec \alpha}$

Left hand side = right hand side

hence proved .

\________________/

#answerwithquality

#BAL