For   $k\in R,$ let the solutions of the equation $cos\left(si{n}^{-1}\left(xcot\left(ta{n}^{-1}\left(cos\left(si{n}^{-1}x\right)\right)\right)\right)\right)$ =, 0 < $\left|x\right|<\frac{1}{\sqrt[]{2}}$  be a and b, where the inverse trigonometric functions take only principal values. If the solutions of the equation x² – bx – 5 = 0 are $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}and\frac{\alpha }{\beta }then\frac{b}{k^2}$  is equal to…………

It's difficult but in some colleges you may can get 

  $\overrightarrow{a}+5\overrightarrow{b}$  is collinear with $\overrightarrow{c}$  

⇒   $\overrightarrow{a}+5\overrightarrow{b}$ = $\overrightarrow{c}$           …(1)

$\overrightarrow{b}+6\overrightarrow{c}$ is collinear with $\overrightarrow{a}$  

⇒   $\overrightarrow{b}+6\overrightarrow{c}=\mu \overrightarrow{a}$               …(2)

From (1) and (2)

  $\overrightarrow{b}+6\overrightarrow{c}=\mu \left(\lambda \overrightarrow{c}-5\overrightarrow{b}\right)$          

-> $\left(1+5\mu \right)\overrightarrow{b}+\left(6-\lambda \mu \right)\overrightarrow{c}=0$

$?\overrightarrow{b}$ and $\overrightarrow{c}$

${\displaystyle \int \frac{\left(sinx-cosx\right)si{n}^{2}x}{tanx\left(si{n}^{3}x+co{s}^{3}x\right)}dx}$

${\displaystyle \int \frac{\left(sinx-cosx\right)sinxcosx}{si{n}^{3}x+co{s}^{3}x}dx}$ , put sin³x + cos³x = t(3 sin²x×cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3}{\displaystyle \int \frac{dt}{t}}$

$=\frac{lnt}{3}+c$

$=\frac{ln\left|si{n}^{3}x+co{s}^{3}x\right|}{3}+c$

$f\left(x\right)=\frac{\left(2^x+2^{-x}\right)tanx\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}}{{\left(7x^2-3x+1\right)}^3}$

$f\left(x\right)=\left(2^x+2^{-x}\right).tanx.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}$

$f\text{'}\left(x\right)=\left(2^x+2^{-x}\right).se{c}^{2}x.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}+tanx.\left(Q\left(x\right)\right)$

$f\text{'}\left(0\right)=2.1\sqrt[]{\frac{\pi }{4}}.1$

$=\sqrt[]{\pi }$