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

[-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).

The given statement is false. \(sin^{-1}?x\) is not same as \((sin?x)^{-1}\). \(sin^{-1}?x\) is an inverse trigonometric function whereas \((sin?x)^{-1}\) is just reciprocal of sin?x i.e. \(sin?x=\frac{1}{sin?x}\).

From ?ABC, we get

tan^-1?(\(\frac{4}{5}\))=cos^-1?(\(\frac{5}{\sqrt{41}}\))
cos?(tan^-1?(\(\frac{4}{5}\))=cos?(cos^-1?(\(\frac{5}{\sqrt{41}}\)))
=\(\frac{5}{\sqrt{41}}\)

Let sec^-1?x=y
?x=sec?y
?x=\(\sqrt{1+tan^2?y}\)
?x²-1=tan²?y
&re4;y=tan^-1?\(\sqrt{x^2-1}\)=sec^-1?x.

There are 2 curves.

The green curve is graph of y = cosx
The red curve is graph for y = cos^-1x
This curve origin from some point before ?/3 and approaches to infinity in both positive y axis by intersecting at a point near 1.5 in y axis.

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?\((\frac{\sqrt{3}}{2})=\frac{?}{2}+\frac{?}{6}=\frac{3?+?}{6}=\frac{2?}{3}\).

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