Question

sqrtx^2 − a^2/x (a > 0), let x = a sec(θ) 0<θ<π/2, and simplify the result.

Answer 1

\[\frac{\sqrt{x^2-a^2}}{x}=\frac{\sqrt{(asec(\theta))^2-a^2}}{asec(\theta)}=\frac{\sqrt{a^2\sec^2(\theta)-a^2}}{asec(\theta)}\]
\[\sqrt{a^2}\frac{\sqrt{\sec^2(\theta)-1}}{asec(\theta)}=a \frac{\sqrt{\sec^2(\theta)-1}}{asec(\theta)}=\frac{\sqrt{\sec^2(\theta)-1}}{\sec(\theta)}\]

but remember
\[\sin^2x+\cos^2x=1\]

so
\[\frac{\sin^2x+\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}\] also holds

\[\tan^2x+1=\sec^2x => \sec^2x-1=\tan^2x\]

so we can write:
\[\frac{\sqrt{\tan^2(\theta)}}{\sec(\theta)}=\frac{\tan(\theta)}{\sec(\theta)}\]
you can write in terms of sin and cos to see if anything else can cancel
\[\tan(\theta) \div \sec(\theta)=\frac{\sin(\theta)}{\cos(\theta)} \div \frac{1}{\cos(\theta)}\]
\[=\frac{\sin(\theta)}{\cos(\theta)}*\cos(\theta)=\sin(\theta)\]

Answer 2

cool?