10. If sin 0 = then prove that (sec @ + tan 0) =
V 6-a
bta
b​

1 thought on “10. If sin 0 = then prove that (sec @ + tan 0) =<br />V 6-a<br />bta<br />b​”

1. Answer:

   Given:

   [tex]sin\,\theta=\frac{a}{b}[/tex]

   [tex]To \: show: sec\,\theta\+tan\,\theta=\sqrt{\frac{b+a}{b-a}}[/tex]

   Using Trigonometric identity,

   [tex]cos\,\theta=\sqrt{1-sin^2\,\theta}[/tex]

   [tex]cos\,\theta=\sqrt{1-(\frac{a}{b})^2}[/tex]

   [tex]cos\,\theta=\sqrt{(\frac{b^2-a^2}{b^2}}[/tex]

   [tex]cos\,\theta=\frac{\sqrt{b^2-a^2}}{b}[/tex]

   [tex]\implies sec\,\theta=\frac{1}{cos\,\theta}=\frac{b}{\sqrt{b^2-a^2}}[/tex]

   [tex]\implies tan\,\theta=\frac{sin\,\theta}{cos\,\theta}=\frac{a}{\sqrt{b^2-a^2}}[/tex]

   Now,

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

   [tex]

   =\frac{b}{\sqrt{b^2-a^2}}+\frac{a}{\sqrt{b^2-a^2}}[/tex]

   [tex]=\frac{b+a}{\sqrt{(b-a)(b+a)}}[/tex]

   [tex]=\frac{\sqrt{b+a}\sqrt{b+a}}{\sqrt{b-a}\sqrt{b+a}}[/tex]

   [tex]=\frac{\sqrt{b+a}}{\sqrt{b-a}}[/tex]

   [tex]=\sqrt{\frac{b+a}{b-a}}[/tex]

   Hence Proved

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