How do you use trigonometric substitution to write the algebraic expression #sqrt(x^2-4)# as a trigonometric function of #theta# where #0<theta<pi/2# and #x=2sectheta#?
1 Answer
Jul 29, 2017
Explanation:
#•color(white)(x)tan^2x=sec^2x-1#
#"substitute "x=2sectheta" into "sqrt(x^2-4)#
#rArrsqrt(4sec^2theta-4)#
#=sqrt(4(sec^2theta-1)#
#=sqrt(4tan^2theta)#
#=2tanthetato0 < theta < pi/2#
