How do you evaluate tan(arccos(-9/10))tan(arccos(−910))?

1 Answer
maganbhai P.
Mar 25, 2018

-sqrt19/9.−√199.

Explanation:

We have, color(red)((1)cos^-1(-x)=pi-cos^-1x(1)cos−1(−x)=π−cos−1x
color(red)((2)tan(pi-theta)=-tantheta(2)tan(π−θ)=−tanθ
color(red)((3)cos^-1x=tan^-1((sqrt(1-x^2)/x)(3)cos−1x=tan−1((√1−x2x)
color(red)((4)tan(tan^-1x)=x(4)tan(tan−1x)=x

tan(cos^-1(-9/10))=tan(pi-cos^-1(9/10)).totan(cos−1(−910))=tan(π−cos−1(910)).→Apply(1)
=-tan(cos^-1(9/10))...........toApply(2)
=-tan(tan^-1(sqrt(1-(9/10)^2)/(9/10)))...toApply(3)
=-tan(tan^-1(sqrt(1-(81/100))/(9/10)))
=-tan(tan^-1((sqrt(100-81)/10)/(9/10)))
=-tan(tan^-1(sqrt19/9))

=-sqrt19/9....to Apply(4)

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