How to simplify tan⁡(sec−1(u))?

Question

ExcendLicceneixi · Accepted Answer

The sides of a triangle can be thought of as the trigonometric functions of an angle, and the inverse trigonometric functions of an angle are functions of a ratio of that triangle that return an angle.
Let there be a right triangle with vertical leg

y

, horizontal leg

x

and hypotenuse

r

.
Let an angle

θ

be defined as the angle formed from the intersection of

x

and

r

measured counter-clockwise.

\sec (θ) = \frac{1}{\cos (θ)} = \frac{r}{x}

so

\sec^{- 1} (u) = \sec^{- 1} (\frac{r}{x}) = θ

The problem now becomes

\tan (θ)

.
We know that

Undefined control sequence \cancel

=

\frac{y}{x}

y1i2de1o64w · Accepted Answer

Let sec−1(u) be equal to some θ.Thus,sec⁡θ=uSquaring the two sides, we obtain,sec2θ=u2tan2θ+1=u2    [we know,tan2θ+1=sec2θ]Thus,tan⁡θ=±u2−1Getting back to the original question,we have,tan⁡(sec−1(u)) which on simplification gives,tan⁡θ which is equal to ±u2−1

How to simplify Tan(sec^-1(u))?

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