Question

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

Answer 1

Please see the attachment

Answer 2

Question :-

Prove that ,

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

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)

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

Left hand side = right hand side

hence proved .

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