How do you evaluate # sin(tan^-1(7/9))#?

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Answer 1 · 2017-01-04T06:31:11

#sin(tan^(-1)(7/9))=7/sqrt130#

Let #theta=tan^(-1)(7/9)#

then #tantheta=7/9# or #cottheta=9/7#

and hence #sin(tan^(-1)(7/9))#

= #sintheta=1/csctheta#

= #1/sqrt(csc^2theta)#

= #1/sqrt(1+cot^2theta)#

= #1/sqrt(1+(9/7)^2)#

= #1/sqrt(1+81/49)=1/sqrt(130/49#

= #7/sqrt130#