Question

if sin thitha = a/b then prove that (sec thitha + tan thitha ) = root b+a/b-a
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Answer 1

sin $\theta$  = $\frac{a}{b}$    ,  to prove (sec$\theta$+tan$\theta$) = $\sqrt{\frac{b+a}{b-a} }$

Step-by-step explanation:

Given, sin $\theta$  = $\frac{a}{b}$  

=> opposite side = a  and hypotenuse = b

then by pythagorous theorem ,

=> b² = a² + x²    , where x is the adjacent side

so, x = $\sqrt{b^2-a^2}$

=> sec$\theta$+tan$\theta$  = $\frac{b}{\sqrt{b^2-a^2}}$ + $\frac{a}{\sqrt{b^2-a^2}}$       , as sec$\theta$ = hyp/adj  & tan$\theta$ = opp/adj

=> sec$\theta$+tan$\theta$  = $\frac{b+a}{\sqrt{b^2-a^2}}$

=> sec$\theta$+tan$\theta$  = $\frac{b+a}{\sqrt{(b+a)(b-a)}}$

=> sec$\theta$+tan$\theta$  = $\frac{\sqrt{b+a}*\sqrt{b+a} }{\sqrt{b+a}*\sqrt{b-a}}$

=> sec$\theta$+tan$\theta$  = $\frac{\sqrt{b+a} }{\sqrt{b-a}}$

=> sec$\theta$+tan$\theta$  = $\sqrt{\frac{b+a}{b-a} }$

and hence we arrived at the equation, that we wanted to prove.

Hence proved.