Question

prove 1-sintheta/1+sintheta =1+2tansquare theta-2tantheta sectheta​

Answer 1

Question to prove :-

${ \dfrac{1 - \sin \theta}{1 + \sin \theta} = 1 + 2 \tan ^{2} \theta - 2 \tan \theta . \sec \theta}$

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Solution :-

» Considering RHS

$\sf{ : \implies 1 + 2 \tan ^{2} \theta - 2 \tan \theta . \sec \theta}$

$\sf{ : \implies 1 + \tan ^{2} \theta + \tan ^{2} \theta - 2 \tan \theta . \sec \theta}$

» Apply identity 1 + tan²θ= sec²θ

$\sf{ : \implies \sec ^{2} \theta + \tan ^{2} \theta - 2 \tan \theta . \sec \theta}$

» Apply identity A²+B²-2AB=(A-B)²

$\sf{ : \implies (\sec \theta - \tan\theta )^{2} }$

» Apply identity sec θ = 1/cos θ

$\sf{ : \implies \bigg(\dfrac{1}{ \cos \theta} - \tan\theta \bigg)^{2} }$

» Apply identity tan θ = sinθ / cosθ

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

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

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

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

» Apply formula cos²θ =1-sin²θ

$\sf{ : \implies \dfrac{ (1 - \sin \theta) (1 - \sin \theta) }{ 1 - \sin ^{2} \theta } }$

$\sf{ : \implies \dfrac{ (1 - \sin \theta) (1 - \sin \theta) }{ (1) ^{2} - \sin ^{2} \theta } }$

» Apply identity A²-B²=(A+B)(A-B)

$\sf{ : \implies \dfrac{ (1 - \sin \theta) (1 - \sin \theta) }{ (1 + \sin \theta)(1 - \sin \theta) } }$

$\sf{ : \implies \dfrac{ \cancel{ (1 - \sin \theta)} (1 - \sin \theta) }{ (1 + \sin \theta) \cancel{(1 - \sin \theta)} } }$

$\sf{ : \implies \dfrac{ (1 - \sin \theta) }{ (1 + \sin \theta) } }$

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

LHS = RHS

Hence proved