Let \(sec^{-1}?(\frac{2}{\sqrt{3}})\)=y
sec? y=\(\frac{2}{\sqrt{3}}\)
sec? y=sec \(?\frac{?}{6}\)
?y=\(\frac{?}{6}\)

We know, sin x = sin(π ? x)
So, sin 6 = sin(π ? 6)
= sin(2π + 6)
= sin(3π ? 6)
= sin(-π ? 6)
= sin(-2π ? 6)
= sin(-3π ? 6)
So, sin^-1(sin 6) = sin^-1(sin (-2π + 6))
= -2π + 6

Using formula sin^-1?x+sin^-1?y=sin^-1?\({x \sqrt{1-y^2}+y \sqrt{1-x^2}}\), we get
sin^-1?(\(\frac{3}{5}\))+sin^-1?(\(\frac{4}{5}\))=sin^-1\(\Big\{ \frac{3}{5} \sqrt{1-(\frac{4}{5})^2}+\frac{4}{5} \sqrt{1-(\frac{3}{5})^2} \Big\}\)
=sin^-1?(\(\frac{3}{5}\)�\(\frac{3}{5}\)+\(\frac{4}{5}\)�\(\frac{4}{5}\))=sin^-1?\((\frac{25}{25})=\frac{?}{2}\)
&re4; sin^-1?(\(\frac{3}{5}\))+sin^-1?(\(\frac{4}{5}\))+cos^-1?\((\frac{\sqrt{3}}{2})=\frac{?}{2}+\frac{?}{6}=\frac{3?+?}{6}=\frac{2?}{3}\).

\(sin^{-1}?(sin?x)\)=x, x∈\([-\frac{?}{2},\frac{?}{2}]\)
&re4;\(sin^{-1} (sin? \frac{4?}{3})=sin^{-1}?(sin?(?+\frac{?}{3}))=sin^{-1}?(sin?(\frac{-?}{3}))= -\frac{?}{3}\).

Let, &ta; = cos^-1(-x)
So, 0 ≤ &ta; ≤ π
=> -x = cos&ta;
=> x = -cos&ta;
=> x = cos(-&ta;)
Also, -π ≤ -&ta; ≤ 0
So, 0 ≤ π -&ta; ≤ π
=> -&ta; = cos^-1(x)
=> &ta; = -cos^-1(x)
So, cos^-1(x) = π ? &ta;
&ta; = π ? cos^-1(x)
=> cos^-1(-x) = π ? cos^-1(x)

The formula sin^-1?x-sin^-1?y=sin^-1?{\(x\sqrt{1+y^2}+y\sqrt{1+x^2}\)} is incorrect. The correct formula is sin^-1?x-sin^-1?y=sin^-1?{\(x\sqrt{1+y^2}-y \sqrt{1-x^2}\)}.

There are 2 curves.

The black curve is graph of y = cotx
The red curve is graph for y = cot^-1x
This curve does not pass through origin but approaches to infinity in direction of x axis only.
The part of curve that lies in (x, y) coordinate gradually meets to x-axis.
This graph lies above +x axis and ?x axis.

[-1, 1] is domain for sin^-1?x.
The domain for cot^-1?x is (-∞,∞).
The domain for tan^-1?x is (-∞,∞).
The domain for sec^-1?x is (-∞,-1]∪[1,∞).

sin^-1sin(6 π/7)
Now, sin(6 π/7) = sin(π – 6 π/7)
= sin(2π + 6 π/7) = sin(π/7)
= sin(3π – 6 π/7) = sin(20π/7)
= sin(-π – 6 π/7) = sin(-15π/7)
= sin(-2π + 6 π/7) = sin(-8π/7)
= sin(-3π – 6 π/7) = sin(-27π/7)
Therefore, sin^-1sin(6 π/7) = sin^-1(π/7).

sin^-1?x+cos^-1?x=\(\frac{?}{2}\); x?[-1,1]