Maths

  $\frac{sin\theta sin\theta }{cos\theta }+\frac{sin\theta }{cos\theta }=2sin\theta cos\theta$

$\Rightarrow \frac{sin\theta }{cos\theta }\left[sin\theta +1\right]=2sin\theta cos\theta$            

$\therefore$ sinθ = 0 and 1 + sin θ = 2 cos² θ = 2 – 2 sin² θ ………….(i)

θ = np

θ = -p, p, 0

From (i), 2 sin² θ + sin θ - 1 = 0

(2 sin θ - 1) (sin θ + 1) = 0

sin θ = -1, $\frac{1}{2}$  

$\theta =\frac{3\pi }{2},\theta =\frac{\pi }{6},\frac{5\pi }{6}$           

$\theta =\frac{3\pi }{2}$ is rejected.

T = cos (-2p) + cos 2p + cos θ + cos $\frac{\pi }{3}+cos\frac{5\pi }{3}=4$  

$\therefore$ T + n(s) = 4 + 5 = 9

$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\frac{2\left(a+b+c\right)}{24}=\frac{c}{5}=\frac{a}{4}=\frac{b}{3}$

$r=\frac{\Delta }{S}=\frac{6k^2}{6k}=k$

$R=\frac{5k}{2}\Rightarrow \frac{R}{r}=\frac{5}{2}$

  $l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{e^{cosx}sinxdx}{\left(1+co{s}^{2}x\right)\left(e^{cosx}+e^{-cosx}\right)}}$ ${\displaystyle \underset{}{\overset{}{}}\frac{{}^{}}{{}^{}{}^{}{}^{}}}$

$2l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{sindx}{1+co{s}^{2}x}=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{sinxdx}{1+co{s}^{2}x}}}$

$l={\displaystyle \underset{1}{\overset{0}{\int }}\frac{-dt}{1+t^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}=\frac{\pi }{4}}}$

$\frac{x^2-5x+6}{x^2-9}\ge -1\&\frac{x^2-5x+6}{x^2-9}\le 1$

$\frac{2x+1}{x+3}\ge 0,x\ne -3\&\frac{1}{x+3}\ge 0,x\ne -3\Rightarrow x>-3$

$x\in \left[-\frac{1}{2},\infty \right]...........(i)$

$x^2-3x+2>0and{x}^2-3x+1\ne 0$

(x – 2) (x – 1) > 0 and $x\ne \frac{3\pm \sqrt[]{5}}{2}$

$x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

From (i) and (ii)   $x\in \left[-\frac{1}{2},1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

4x – 3y + k₂ = 0

$2r=\frac{40}{5}=8\Rightarrow r=4$

${\left(x-1\right)}^2+{\left(y+2\right)}^2=16$

n = 33, p = success, q = failure

3P (x = 0) = P (x = 1)

${\Rightarrow }^{33}{C}_0{p}^0{q}^{33}{=}^{33}{C}_{1}p{q}^{32}$            

$\Rightarrow p=\frac{1}{12},q=\frac{11}{12}\Rightarrow \frac{q}{p}=11$          

………. (i)

Subtracting, (ii) – (i), we get 1320

$\left|\widehat{a}\right|=1,\left|\widehat{b}\right|=1$

${\left|\widehat{b}\right|}^2={\left|\overrightarrow{c}+2\left(\overrightarrow{c}*\widehat{a}\right)\right|}^2$

$\Rightarrow 1={\left|\overrightarrow{c}\right|}^2+4{\left|\overrightarrow{c}\right|}^2{\left(\frac{\sqrt[]{3}-1}{2\sqrt[]{2}}\right)}^2$

${\left|\overrightarrow{c}\right|}^2\left[1+4*\frac{{\left(\sqrt[]{3}-1\right)}^2}{8}\right]={\left|\overrightarrow{c}\right|}^2\left(3-\sqrt[]{3}\right)$

$\therefore {\left|\overrightarrow{c}\right|}^2=\frac{1}{3-\sqrt[]{3}}=\frac{3+\sqrt[]{3}}{6}$

$\Rightarrow {\left|6\overrightarrow{c}\right|}^2=6\left(3+\sqrt[]{3}\right)$

$\lambda x-2y=\mu ........(i)and\frac{x^2}{\left(\frac{b^2}{a^2}\right)}-\frac{y^2}{b^2}=1.........(ii)$

Let (x₁, y₁) be a point on the curve equation of tangent of eq (ii)

  $y=mx\pm \sqrt[]{\frac{b^2}{a^2}{m}^2-b^2}..........(iii)$

From (i) and (ii) are identical

$\therefore m=\frac{\lambda }{2}and\frac{b^2}{a^2}{m}^2-b^2=\frac{{\mu }^2}{4}$               

$\Rightarrow \frac{b^2}{a^2}*\frac{{\lambda }^2}{4}-b^2=\frac{{\mu }^2}{4}$

$\Rightarrow {\left(\frac{\lambda }{a}\right)}^2-{\left(\frac{\mu }{b}\right)}^2=4$

  $y\frac{dx}{dy}=2x+y^3\left(y+1\right)e^y$               

    $\frac{dx}{dy}-\frac{2}{y}x=y^3\left(1+y\right)e^y..........(i)$

Equation (i) is in linear form, so I.F. = y^-2

$\begin{array}{l}\therefore x{y}^{-2}=y{e}^y+c\Rightarrow 0=e+c\Rightarrow c=-e\\ \therefore x=e^3\left(e^e-1\right)\end{array}$