Respuesta :

Answer:

Hi mate!

To prove:-

[tex] \bf \sqrt{ \frac{1 + \sin( \theta)}{1 - \sin( \theta)} } = \sec( \theta) - \tan( \theta) [/tex]

Step-by-step explanation:

By taking LHS,

[tex] \maltese \: \: \large \bf \sqrt{ \frac{1 + \sin( \theta)}{1 - \sin( \theta)} } [/tex]

By rationalisation method,

[tex] : \longrightarrow \bf \sqrt{ \frac{1 + \sin( \theta)}{1 - \sin( \theta)} \times \frac{1 + \sin( \theta)}{1 + \sin( \theta)} } \\ \\ : \longrightarrow \frac{ {( \sqrt{1 + \sin\theta }) }^{2} }{( \sqrt{1 - \sin( \theta)})(\sqrt{1 + \sin( \theta)}) } \\ \\ : \longrightarrow \frac{ {( \sqrt{1 + \sin\theta }) }^{2} }{ \sqrt{ {1}^{2} - { \sin}^{2} \theta } } \: \\ \\ : \longrightarrow \: \frac{1 + \sin \theta}{ \cos \theta } \\ \\ \longrightarrow \: \frac{1}{ \cos \theta } + \frac{ \sin \theta}{ \cos \theta} [/tex]

[tex] \: \: \: \: \: \bigg\lgroup \because \: \frac{1}{ \cos \theta }\: = \sec \theta \: \\ \: \: \: \: \: \: \: \: \: \: \: \frac{ \sin \theta}{ \cos \theta} = \tan \theta \bigg \rgroup[/tex]

[tex] \implies \boxed{\pink{ \sec \theta + \tan \theta }}[/tex]

Now taking RHS

[tex] \sf \sec \theta + \tan \theta [/tex]

[tex]\implies \: \: \frac{1}{ \cos \theta } + \frac{ \sin \theta}{ \cos \theta} \\ \\ \implies \frac{1 + \sin \theta }{ \cos \theta } [/tex]

Hence, proved

Answer:

[tex]\large{\pink{\underline\textsf{To Prove :-}}}[/tex]

[tex] \sqrt{ \frac{1 + \sin \theta }{1 - \sin \theta } } = \sec \theta + \tan \theta[/tex]

[tex] \huge{\orange{\underline{\textsf{Solution :-}}}}[/tex]

[tex] \sf \sqrt{ \frac{1 + \sin \theta }{1 - \sin \theta } } [/tex]

  • First we have to rationalize it to remove the square root
  • Now we will multiply √1+sinθ both sides

[tex] \sqrt{ \frac{1 + \sin \theta }{1 - \sin \theta } } \times \sqrt{ \frac{1 + \sin \theta }{1 - \sin \theta } } \\ \sqrt{ \frac{ {(1 + \sin \theta) }^{2} }{(1 + \sin \theta)(1 - \sin \theta) } } \\ \sqrt{ \frac{ {(1 + \sin \theta) }^{2} }{1 - { { \sin}^{2} \theta } } }[/tex]

  • We know that sin²θ + cos²θ = 1
  • so cos²θ = 1 - sin²θ

[tex] \sqrt{ \frac{( {1 + \sin \theta })^{2} }{ { \cos}^{2} \theta} } \\ \frac{1 + \sin \theta }{ \cos \theta } \\ \frac{1}{ \cos \theta} + \frac{ \sin \theta }{ \cos \theta } [/tex]

[tex]\sf{\red { \sec \theta + \tan \theta}}[/tex]

[tex] \sf \huge \purple {HENCE \: \: PROVED! }[/tex]

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