Inverse Trigonometric Functions

Kindly consider the following figure

Kindly consider the following figure

2π - (sin? ¹ (4/5) + sin? ¹ (5/13) + sin? ¹ (16/65)
= 2π - (tan? ¹ (4/3) + tan? ¹ (5/12) + tan? ¹ (16/63)
= 2π - (tan? ¹ (63/16) + tan? ¹ (16/63)
= 2π - π/2 = 3π/2

f (x)=sin (|x|+5)/ (x²+1)
For domain:
-1 ≤ (|x|+5)/ (x²+1) ≤ 1
Since |x|+5 and x²+1 is always positive
So (|x|+5)/ (x²+1) ≥ 0 ∀x∈R
So for domain:
(|x|+5)/ (x²+1) ≤ 1
⇒ |x|+5 ≤ x²+1
⇒ 0 ≤ x²-|x|-4
⇒ 0 ≤ (|x|- (1+√17)/2) (|x|- (1-√17)/2)
⇒ |x| ≥ (1+√17)/2 or |x|≤ (1-√17)/2 (Rejected)
⇒ x∈ (-∞, - (1+√17)/2] ∪ [ (1+√17)/2, ∞)
So, a = (1+√17)/2

y = (8²?-8?²?)/(8²?+8?²?) ⇒ (1+y)/(1-y) = 8? ⇒ 4x = log?((1+y)/(1-y))
x = (1/4)log?((1+y)/(1-y)), f?¹(x) = (1/4)log?((1+x)/(1-x))

Given $ta{n}^{-1}a+t{a}^{-1}b=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ ${}^{}{}^{}\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$

$ta{n}^{-1}\left(\frac{a+b}{1-ab}\right)=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

$\frac{a+b}{1-ab}=1...........\left(i\right)$

OR a + b = 1 – ab .(ii)

Now, $\left(a+b\right)-\left(\frac{a^2+b^2}{2}\right)+\left(\frac{a^3+b^3}{3}\right)-\left(\frac{a^4+b^4}{4}\right)+........$ $\frac{{}^{}{}^{}}{}\frac{{}^{}{}^{}}{}\frac{{}^{}{}^{}}{}$

$\left(a-\frac{a^2}{2}+\frac{a^3}{3}+\frac{a^3}{3}-\frac{a^4}{4}+......\right)+\left(b-\frac{b^2}{2}+\frac{b^3}{3}-\frac{b^4}{4}+.......\right)$

log (1 + a) + log(1 + b) = log (1 + a) (1 + b) = log {1 + a + b + ab} = log_e2

  $g\left(2\right)=\underset{x\to 2}{lim}g\left(x\right)=\underset{x\to 2}{lim}\frac{x^2-x-2}{2x^2-x-6}=\underset{x\to 2}{lim}\frac{\left(x-2\right)\left(x+1\right)}{2x\left(x-2\right)+3\left(x-2\right)}$         

$Nowfog=f\left(g\left(x\right)\right)=si{n}^{-1}g\left(x\right)=si{n}^{-1}\left(\frac{x^2-x-2}{2x^2-x-6}\right)$

$\frac{3x^2-2x-8}{2x^2-x-6}\ge 0\&\frac{-x^2+4}{2x^2-x-6}\le 0$

On solving we get  $x\in \left(-\infty ,-2\right)\cup \left[\frac{-4}{3},\infty \right)$

As x = 2 also lies in domain since g(2) $=\underset{x\to 2}{lim}g\left(x\right)$