Maths

 $x,y\in \left(0,\pi \right)$

cos x + cos y – cos (x + y) = $\frac{3}{2}$

$2cos\frac{x+y}{2}.cos\frac{x-y}{2}-\left(2co{s}^2\frac{x+y}{2}-1\right)=\frac{3}{2}$

$\Rightarrow 2co{s}^2\frac{x+y}{2}-2cos\frac{x-y}{2}.cos\frac{x+y}{2}+\frac{1}{2}=0$

As,

$x,y\in \left(0,\pi \right)$

$\frac{-\pi }{2}<\frac{x-y}{2}<\frac{\pi }{2}$

$sin\frac{x-y}{2}=0\Rightarrow \frac{x-y}{2}=0\Rightarrow x=y$

then equation : cos x + cos y – cos (x + y) = $\frac{3}{2}$

$cosx=\frac{2\pm \sqrt[]{4-4}}{4}=\frac{1}{2}$

$x=y=\frac{\pi }{3}\Rightarrow sinx+cosy=\frac{\sqrt[]{3}}{2}+\frac{1}{2}=\frac{\sqrt[]{3}+1}{2}$

$z^2+\alpha z+\beta =0,\alpha ,\beta \in R$

roots : 1 – 2i, 1 + 2i

Sum of roots = 2 = -α and product of roots = 5 = β

α - β = -2 - 5 = -7

As per questions

$\frac{dy}{dx}=\frac{x^2-4x+y+8}{x-2}$           

$\frac{dy}{dx}=\frac{{\left(x-2\right)}^2+\left(y+4\right)}{\left(x-2\right)}$           

$\frac{dy}{dx}=\left(x-2\right)+\frac{y+4}{x-2}$                       .(i)

Let $\frac{y+4}{x-2}=t$  

(y + 4) = t(x – 2)

Putting in equation (i)

$\left(x-2\right)\frac{dt}{dx}+t=\left(x-2\right)+t$        

    $\frac{dt}{dx}=1$        

dt = dx

Integrating on both the sides t = x + c

$\frac{y+4}{x-2}=x+c$  

Passing through origin C = -2

  $\therefore$         equation of curve $\frac{y+4}{x-2}=x-2$

Combined equation of pair of lines OP and OQ is

$x^2+2y^2=2{\left(x+y\right)}^2$

$\Rightarrow x^2+4xy=0\Rightarrow x\left(x+4y\right)=0\Rightarrow \{\begin{array}{l}x=0\left(lineOP\right)\\ y=-\frac{x}{4}\left(lineOQ\right)\end{array}$

$tan\left(90°+\theta \right)=-\frac{1}{4}$

$-cot\theta =-\frac{1}{4}\Rightarrow tan\theta =4$

$\theta =ta{n}^{-1}4=co{t}^{-1}\frac{1}{4}=\frac{\pi }{2}-ta{n}^{-1}\frac{1}{4}$

$\angle POQ=\pi -\theta =\frac{\pi }{2}+ta{n}^{-1}\frac{1}{4}$

$l_n={\displaystyle \underset{\pi /4}{\overset{\pi /2}{\int }}co{t}^{n}xdx=}{\displaystyle \underset{\pi /4}{\overset{\pi /2}{\int }}co{t}^{n-2}x\left(cose{c}^{2}x-1\right)dx}$

$={\left[\frac{-co{t}^{n-1}x}{n-1}\right]}_{\pi /4}^{\pi /2}-l_{n-2}=\frac{1}{n-1}-l_{n-2}$

$l_n+l_{n-2}=\frac{1}{n-1}$

$\begin{array}{l}n=4\Rightarrow l_4+l_2=\frac{1}{3}\\ n=5\Rightarrow l_5+l_3=\frac{1}{4}\\ n=6\Rightarrow l_6+l_4=\frac{1}{5}\end{array}\}\Rightarrow \frac{1}{l_2+l_4},\frac{1}{l_3+l_5},\frac{1}{l_4+l_6}$ are in A.P.

2x + 3y + 2z = 9 . (i)

3x + 2y + 2z = 9     . (ii)

x – y + 4z = 8          . (iii)

(ii) – (i) ⇒ x = y

Then (iii) ⇒ z = 2

(i) ⇒ 5x + 4 = 9 ⇒ (x = 1, y = 1, z = 2) unique solution

${\alpha }^2-6\alpha -2=0,{\beta }^2-6\beta -2=0$

${\alpha }^2-2=6\alpha ,{\beta }^2-2=6\beta$

$\frac{a_{10}-2a_8}{3a_9}=\frac{{\alpha }^{10}-{\beta }^{10}-2\left({\alpha }^8-{\beta }^8\right)}{3\left({\alpha }^9-{\beta }^9\right)}=\frac{{\alpha }^8\left({\alpha }^2-2\right)-{\beta }^8\left({\beta }^2-2\right)}{3\left({\alpha }^9-{\beta }^9\right)}$

$\frac{{\alpha }^8.6\alpha -{\beta }^8.6\beta }{3\left({\alpha }^9-{\beta }^9\right)}=2$

Yes. Determinants can be calculated for any square matrix of n-order, and it is done by expansion of rows and columns. Even in higher dimensions, their job is to define hyper volumes and transformations.

In such cases, the value of determinant turns out to be zero. this is because swapping the values changes the magnitude into the opposite sign (fundamental property of determinants), which results in the final answer being zero.

Yes. Determinants can be fractions and irrational numbers depending on the values of the matrix. This is because the values are calculated as sums and products of the numbers in the matrix which can turn out to be any type of integer.