How do you use trigonometric substitution to write the algebraic expression #sqrt(x^2-9)# as a trigonometric function of #theta# where #0<theta<pi/2# and #x=3sectheta#?

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Answer 1 · 2017-02-13T12:12:03

The desired function is #3tantheta#

You have already given hint for the solution.

If we put #x=3sectheta#, then #sqrt(x^2-9)#

= #sqrt(9sec^2theta-9)#

= #sqrt(9(sec^2theta-1))#

= #sqrt(9tan^2theta)#

= #3tantheta#

Hence, the desired function is #3tantheta#