How do you simplify Tan(sec^-1(u))tan(sec1(u))?

2 Answers
Apr 27, 2018

y/xyx

Explanation:

Trigonometric functions of an angle can be thought of as ratios of the sides of a triangle and its inverse functions are functions of a ratio of that triangle that return an angle.

Let there be a right triangle with vertical leg yy, horizontal leg xx and hypotenuse rr.
Let an angle thetaθ be defined as the angle formed from the intersection of xx and rr measured counter-clockwise.

sec(theta) = 1/cos(theta) = r/xsec(θ)=1cos(θ)=rx
so sec^-1(u)=sec^-1(r/x)=thetasec1(u)=sec1(rx)=θ
The problem now becomes tan(theta)tan(θ).

We know that tan(theta)=sin(theta)/cos(theta)=(y/cancel(r))/(x/cancel(r)) = y/x

Apr 27, 2018

see below

Explanation:

Let sec^-1(u) be equal to some theta.
Thus,
sec theta=u
Squaring both sides,we get,
sec^2 theta=u^2
tan^2theta+1=u^2 [we know,tan^2theta+1=sec^2 theta]
Thus,
tan theta=+-sqrt(u^2-1)

Getting back to the original question,we have,
tan(sec^-1(u)) which on simplification gives,
tan theta which is equal to +-sqrt(u^2-1)