Question

if tan theta=5a-a/20,then sec theta+tan theta=?​

Answer 1

Refer to Attachment!!!!

Thanks!!!!

Answer 2

Answer:

$sec$ θ $+tan$ θ $=\frac{\sqrt{25+a^2}+a }{5}$

Step-by-step explanation:

$tan$ θ $=\frac{5a-a}{20}$

$=\frac{4a}{20} \\\\=\frac{a}{5}$

$1+tan^2$ θ $= sec^2$ θ

$sec$ θ $=\sqrt{1+tan^2}$θ

$sec$ θ $= \sqrt{1+\frac{a^2}{25} }$

$=\sqrt{\frac{25+a^2}{25} }$

$=\frac{\sqrt{25+a^2} }{5}$

$sec$ θ $+tan$ θ $=\frac{\sqrt{25+a^2} }{5} +\frac{a}{5}$

$sec$ θ $+tan$ θ $=\frac{\sqrt{25+a^2}+a }{5}$