Question

`tan(sec^(-1)sqrt((u^2+9)/u))=` ? im not sure how to solve this please help ?

Answer 1

`tan(sec^(-1)(sqrt((u^2+9)/u)))=sqrt((u^2-u+9)/u)`

Explanation:

Let `sec^(-1)(sqrt((u^2+9)/u))=x` then

`rarrsecx=sqrt((u^2+9)/u)`

`rarrtanx=sqrt(sec^2x-1)=sqrt((sqrt((u^2+9)/u))^2-1)`

`rarrtanx=sqrt((u^2+9-u)/u)=sqrt((u^2-u+9)/u)`

`rarrx=tan^(-1)(sqrt((u^2-u+9)/u))=sec^(-1)(sqrt((u^2+9)/u))`

Now, `tan(sec^(-1)(sqrt((u^2+9)/u)))=tan(tan^(-1)(sqrt((u^2-u+9)/u)))=sqrt((u^2-u+9)/u)`

Answer 2

Rule:-`" "color(red)(ul(bar(|color(green)(sec^-1(x/y)=tan^-1(sqrt(x^2-y^2)/y))|`

`tan(sec^(-1)sqrt((u^2+9)/u))`

`=tan(sec^-1(sqrt(u^2+9)/sqrtu))`

`=tan(tan^-1(sqrt((sqrt(u^2+9))^2-(sqrtu)^2)/sqrtu))`

`=tan(tan^-1(sqrt(u^2+9-u)/sqrtu))`

`=sqrt(u^2+9-u)/sqrtu`

`=sqrt(u+9/u-1)`

Hope it helps...
Thank you...

:-)

You can easily find the derivation of the rule I used. Try it.

My this incomplete scratchpad may help you.

Make the inverse functions into trigonometric functions and then solve it.