Question

1-sin theta/1+sin Theta=(sec theta-tan theta)^2
​

Answer 1

Answer:

To prove:-

\frac{1-sin \theta}{1 +sin \theta} = ( \sec \theta - tan \theta)^{2} \\ \\

Solution:-

L.H.S

\frac{1-sin \theta}{1 +sin \theta} \\ = \frac{(1 - \sin \theta)(1 - \sin \theta) }{(1 + \sin \theta ) (1 - \sin \theta )} \\ = \frac{(1 - \sin \theta)^{2} }{ {(1)}^{2} - (\sin \theta)^{2} \ } \\ = \frac{ {(1 - \sin \theta )}^{2} }{1 - \ { \sin }^{2} \theta } \\ = \frac{ {(1 - \sin \theta )}^{2} }{cos^{2} \theta} \\ = ( \frac{1 - \sin \theta }{cos \theta} )^{2} \\ = (\frac{1}{ \cos \theta } - \frac{ \sin \theta}{ \cos\theta } )^{2} \\ = ( \sec\theta - tan\theta)^{2} \\

Therefore,

\frac{1-sin \theta}{1 +sin \theta} = ( \sec \theta - tan \theta)^{2} \\ \\

L.H.S = R.H.S

Hence proved.

Answer 2

[tex]\mathfrak{\huge{Answer:-}}

\\ \bold{ \frac{1 - sinθ}{1 + sinθ} = {(secθ - tanθ)}^{2} } \\ \sqrt{ \frac{1 - sinθ}{1 + sinθ} } = secθ - tanθ \\ \sqrt{ \frac{1 - sinθ}{1 + sinθ} } \times \sqrt{ \frac{1 - sinθ}{1 + sinθ} } = secθ - tanθ\\ \sqrt{ \frac{(1 -sinθ )(1 -sinθ )}{(1 +sin θ)(1 - sinθ)} } =secθ - tanθ \\ \sqrt{ \frac{ {(1 - sinθ)}^{2} }{ {(1)}^{2} - {(sinθ)}^{2}} } = secθ - tanθ\\ \sqrt{ \frac{ {(1 - sinθ)}^{2} }{1 - {sin}^{2} θ} } secθ - tanθ = \\ \sqrt{ \frac{ {(1 -sin θ)}^{2} }{ {cos}^{2}θ } } =secθ - tanθ \\ \frac{1 - sinθ}{cosθ} = secθ - tanθ \\ \frac{1}{cosθ} - \frac{sinθ}{cosθ} = secθ - tanθ \\ secθ - tanθ=secθ - tanθ\\LHS=RHS\\Hence\:proved[/tex]