Question

If sin =a/b show that sec+tan =whole root b+a/b-a

Answer 1

Given:

$Sin\theta = \frac{a}{b}$

$a = bSin\theta$

$a^2 = b^2(1-cos^2\theta)$

$Cos^2\theta = \frac{b^2-a^2}{b^2}$

$Cos\theta = \frac{\sqrt{b^2-a^2} }{b}$

Hence,

$Sec\theta = \frac{1}{Cos\theta} = \frac{b}{\sqrt{b^2-a^2} }$

$Tan\theta = Sin\theta Sec\theta = \frac{a}{\sqrt{b^2-a^2} }$

$Sec\theta+Tan\theta = \frac{1}{\sqrt{b^2-a^2} }(b+a)$

$Sec\theta+Tan\theta = \sqrt{\frac{(b+a)^2}{(b+a)(b-a)} }$

$Sec\theta+Tan\theta = \sqrt{\frac{(b+a)}{(b-a)} }$

Hence Proved