Trigonometric Functions

$$\begin{gathered} \sum _{r=1}^{13}\frac{1}{\sin \left[\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right]\sin \left[\frac{\pi }{4}+\frac{r\pi }{6}\right]} \\ \sum _{r=1}^{13}\frac{\sin \left[\frac{\pi }{4}+r\frac{\pi }{6}\right]-\left[\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right]}{\left(\sin \frac{\pi }{6}\right)\times \sin \left(\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right)\sin \left(\frac{\pi }{4}+\frac{r\pi }{6}\right)} \\ \sum _{r=1}^{13}2\left(\cot \left(\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right)-\cot \left(\frac{\pi }{4}+\frac{r\pi }{6}\right)\right) \\ 2\left[\cot \frac{\pi }{4}-\cot \left(\frac{\pi }{4}+\frac{13\pi }{6}\right)\right] \\ 2\left(1-\cot \frac{5\pi }{12}\right) \\ 2(1-2+\sqrt{3})\Rightarrow (2\sqrt{3}-2) \end{gathered}$$

\(\begin{aligned}
& 2 x^2+(\cos \theta) x-1=0 \\
& \alpha+\beta=\frac{-\cos \theta}{2}; \quad \alpha \beta=-\frac{1}{2} \\
& \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta =\frac{\cos ^2 \theta}{4}+1 \\
& \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{2}{4} \\
& \alpha^4+\beta^4=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2}
\end{aligned}
\)

Maximum when \(\cos \theta=1\)

\(
\begin{aligned}
& M=\left(\frac{1}{4}+1\right)^2-\frac{1}{2} \\
& M=\frac{17}{16}
\end{aligned}
\)

Minimum when \(\cos \theta=0\)

\(
m=1-\frac{1}{2}=\frac{1}{2}
\)

\(\left.\therefore \quad 16(M+m)=16 (\frac{17}{16}+\frac{1}{2}\right)=25\)

Given, \(\cos \theta+1=\sqrt{3} \sin \theta\)

\(2 \cos ^2 \frac{\theta}{2}=\sqrt{3} \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\)

\(2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sqrt{3} \sin \frac{\theta}{2}\right)=0\)

If \(\cos \frac{\theta}{2}=0\)

Now in the interval \((-2 \pi, 2 \pi)\), the possible values are.

\(\theta=-\pi, \pi\)

If \(\cos \frac{\theta}{2}-\sqrt{3} \sin \frac{\theta}{2}=0\)

\(\tan \frac{\theta}{2}=\frac{1}{\sqrt{3}}\)

Now in the interval \((-2 \pi, 2 \pi)\), the possible values are:

\(\theta=-\frac{5 \pi}{3}, \frac{\pi}{3}\)

Therefore,

\(S=\left\{-\pi, \pi,-\frac{5 \pi}{3}, \frac{\pi}{3}\right\}\)

Hence,

\(\sum_{\theta \in S} \theta=-\frac{4 \pi}{3}\)

$$\begin{gathered} \frac{3\cos 36°+5\sin 18°}{5\cos 36°-3\sin 18°} \\ =\frac{\frac{3(\sqrt{5}+1)}{4}+5\left(\frac{\sqrt{5}-1}{4}\right)}{5\left(\frac{\sqrt{5}-1}{4}\right)-3\left(\frac{\sqrt{5}-1}{4}\right)} \end{gathered}$$

$=\frac{8\sqrt{5}-2}{2\sqrt{5}+8}$

$=\frac{4\sqrt{5}-1}{\sqrt{5}+4}\times \frac{\sqrt{5}-4}{\sqrt{5}-4}$

$=\frac{20-16\sqrt{5}-\sqrt{5}+4}{-11}$

$$\begin{gathered} =\frac{17\sqrt{5}-24}{11} \\ \Rightarrow a=17,b=24,c=11 \end{gathered}$$

$a+b+c=52$

$$\begin{gathered} \frac{3\cos 36°+5\sin 18°}{5\cos 36°-3\sin 18°} \\ =\frac{\frac{3(\sqrt{5}+1)}{4}+5\left(\frac{\sqrt{5}-1}{4}\right)}{5\left(\frac{\sqrt{5}-1}{4}\right)-3\left(\frac{\sqrt{5}-1}{4}\right)} \end{gathered}$$

$=\frac{8\sqrt{5}-2}{2\sqrt{5}+8}$

$=\frac{4\sqrt{5}-1}{\sqrt{5}+4}\times \frac{\sqrt{5}-4}{\sqrt{5}-4}$

$=\frac{20-16\sqrt{5}-\sqrt{5}+4}{-11}$

$$\begin{gathered} =\frac{17\sqrt{5}-24}{11} \\ \Rightarrow a=17,b=24,c=11 \end{gathered}$$

$a+b+c=52$

\(\sin x+\sin ^2 x=1\)

\(\Rightarrow \sin x=\cos ^2 x\) and \(\tan x=\cos x\)

Given expression becomes

\(=2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x\)

\(=2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right]\)

\(=2 \sin ^3 x\left[(\sin x+1)^3\right]\)

\(=2\left[\sin ^2 x+\sin x\right]^3\)

\(=2\)

\(f(x)=2x+3\tan x-\pi\)

\(f'(x)=2+3\sec^2x\)

\(\sec^2x>0\),

\(f'(x)>0\)

for every point in the domain.

Therefore \(f(x)\) is strictly increasing on each interval

\(\left(-2\pi,-\frac{3\pi}{2}\right)\),

\(\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right)\),

\(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\),

\(\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\),

\(\left(\frac{3\pi}{2},2\pi\right)\).

Now check sign changes.

\(\left(-2\pi,-\frac{3\pi}{2}\right)\),

\(f(-2\pi)=-5\pi<0\),

\(f(x)\to+\infty\) as \(x\to\left(-\frac{3\pi}{2}\right)^-\).

Exactly one solution.

\(\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right)\),

\(f(x)\to-\infty\) as \(x\to\left(-\frac{3\pi}{2}\right)^+\),

\(f(x)\to+\infty\) as \(x\to\left(-\frac{\pi}{2}\right)^-\).

Exactly one solution.

\(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\),

\(f(x)\to-\infty\) as \(x\to\left(-\frac{\pi}{2}\right)^+\),

\(f(x)\to+\infty\) as \(x\to\left(\frac{\pi}{2}\right)^-\).

Exactly one solution.

\(\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\),

\(f(x)\to-\infty\) as \(x\to\left(\frac{\pi}{2}\right)^+\),

\(f(x)\to+\infty\) as \(x\to\left(\frac{3\pi}{2}\right)^-\).

Exactly one solution.

\(\left(\frac{3\pi}{2},2\pi\right)\),

\(f(x)\to-\infty\) as \(x\to\left(\frac{3\pi}{2}\right)^+\),

\(f(2\pi)=3\pi>0\).

Exactly one solution.

Total number of solutions

\(=1+1+1+1+1=5\).

Given $(\mathrm{x},\mathrm{y})=\left(3\tan \left(\mathrm{\theta }+\frac{\mathrm{\pi }}{3}\right),2\tan \left(\mathrm{\theta }+\frac{\mathrm{\pi }}{6}\right)\right)$

\( \mathrm{x}=3\left(\frac{\tan \theta+\sqrt{3}}{1-\sqrt{3} \tan \theta}\right) \)
\( \mathrm{x}-\sqrt{3} \tan \theta=3 \tan \theta+3 \sqrt{3} \)
\( \tan \theta=\frac{x-3 \sqrt{3}}{3+\sqrt{3} x} \quad \ldots(1)\)

\( 2\left(\frac{\tan \theta+\frac{1}{\sqrt{3}}}{1-\frac{\tan \theta}{\sqrt{3}}}=y\right) \)

\( 2(\sqrt{3} \tan \theta+1)=y(\sqrt{3}-\tan \theta) \ldots(2) \)
using (1) and (2)
\( 2\left(\frac{x-3 \sqrt{3}}{\sqrt{3}+x}+1\right)=y\left(\sqrt{3}-\frac{(x-3 \sqrt{3})}{\sqrt{3}(\sqrt{3}+x)}\right) \)
\( 2 \sqrt{3}(x-3 \sqrt{3}+x+\sqrt{3})=y(3(\sqrt{3}+x)-x+3 \sqrt{3})\)
\( 4 \sqrt{3} x-12=y(2 x+6 \sqrt{3}) \)
\(x y-2 \sqrt{3} x+3 \sqrt{3} y-6=0 \)
\( \Rightarrow \alpha=-2 \sqrt{3}, \beta=3 \sqrt{3}, \gamma=-6\)
\( \alpha^2+\beta^2+\gamma^2=12+27+36=75\)

\(\begin{aligned}
& 2 x^2+(\cos \theta) x-1=0 \\
& \alpha+\beta=\frac{-\cos \theta}{2}; \quad \alpha \beta=-\frac{1}{2} \\
& \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta =\frac{\cos ^2 \theta}{4}+1 \\
& \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{2}{4} \\
& \alpha^4+\beta^4=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2}
\end{aligned}
\)

Maximum when \(\cos \theta=1\)

\(
\begin{aligned}
& M=\left(\frac{1}{4}+1\right)^2-\frac{1}{2} \\
& M=\frac{17}{16}
\end{aligned}
\)

Minimum when \(\cos \theta=0\)

\(
m=1-\frac{1}{2}=\frac{1}{2}
\)

\(\left.\therefore \quad 16(M+m)=16 (\frac{17}{16}+\frac{1}{2}\right)=25\)

Let $S=\sum _{r=1}^{13}\frac{1}{\sin \left(\frac{\pi }{4}+\frac{r\pi }{6}\right)\sin \left(\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right)}$

$=2\sum _{r=1}^{13}\frac{\sin \left(\frac{\pi }{4}+\frac{r\pi }{6}\right)-\left(\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right)}{\sin \left(\frac{\pi }{4}+\frac{\pi }{6}\right)\sin \left(\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right)}$

$=2\sum _{r=1}^{13}\left(\cot \left(\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right)-\cot \left(\frac{\pi }{4}+\frac{\mathrm{r}\pi }{6}\right)\right)$

$=2\left[\cot \left(\frac{\pi }{4}\right)-\cot \left(\frac{\pi }{4}+\frac{13\pi }{6}\right)\right]$

$=2\left[1-\cot \left(\frac{\pi }{4}+2\pi +\frac{\pi }{6}\right)\right]$

$=2\left[1-\cot \left(\frac{5\pi }{12}\right)\right]=2\left(1-2+\sqrt{3}\right)$

$=2\sqrt{3}-2=a\sqrt{3}+b$

$a^2+b^2=8$

Given:

\( f(\theta)=4\left(\sin ^4\left(\frac{7 \pi}{2}-\theta\right)+\sin ^4(11 \pi+\theta)\right) -2\left(\sin ^6\left(\frac{3 \pi}{2}-\theta\right)+\sin ^6(9 \pi-\theta)\right)\)

\(\Rightarrow f(\theta)=4\left(\cos ^4(\theta)+\sin ^4(\theta)\right)-2\left(\cos ^6 \theta+\sin ^6 \theta\right)\)

\(\Rightarrow f(\theta)=4\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)-2\left(1-3 \sin ^2 \theta \cos ^2 \theta\right)\)

\(\Rightarrow f(\theta)=2-2 \sin ^2 \theta \cos ^2 \theta\)

\(\Rightarrow f(\theta)=2-\frac{\sin ^2(2 \theta)}{2}\)

So, \(\alpha=f(\theta)_{\max }=2\),

\(\beta=f(\theta)_{\min }=\frac{3}{2}\)

\(\Rightarrow \alpha+2 \beta=5\)

$\sin x=\frac{-3}{5},\pi <x<\frac{3\pi }{2}$

$\Rightarrow \tan x=\frac{3}{4}$ and $\cos x=-\frac{4}{5}$

$80\left({\tan }^{2}x-\cos x\right)=80\left(\frac{9}{16}+\frac{4}{5}\right)=45+64=109$

$10{\sin }^4\theta +15{\cos }^4\theta =6$

\(\Rightarrow 10\left[1-2 \sin ^2 \theta \cos ^2 \theta\right]+5 \cos ^4 \theta=6\)

\(\Rightarrow 25 \cos ^4 \theta-20 \cos ^2 \theta+4=0\)

\(\Rightarrow\left(5 \cos ^2 \theta-2\right)^2=0\)

$\Rightarrow {\cos }^2\theta =\frac{2}{5}$ and ${\sin }^2\theta =\frac{3}{5}$

$$\begin{gathered} \Rightarrow \frac{27{cosec}^6\theta +8{\sec }^6\theta }{16{\sec }^8\theta } \\ =\frac{27·{\left(\frac{5}{3}\right)}^3+8·{\left(\frac{5}{2}\right)}^3}{16·{\left(\frac{5}{4}\right)}^4} \\ =\frac{2}{5} \end{gathered}$$

$\sqrt{3}{\mathrm{cosec}}^2\theta -2(\sqrt{3}-1)\mathrm{cosec}\theta -4=0$

Let $x = \csc \theta$.

$\sqrt{3} x^2 - 2(\sqrt{3} - 1) x - 4 = 0$

$\sqrt{3}x(x - 2) + 2(x - 2) = 0$

$(\sqrt{3}x + 2)(x - 2) = 0$

So, the solutions for $x$ are: \(x=2 \) or \(x=-\frac{2}{\sqrt{3}}\)

Substituting $x=\mathrm{cosec}\theta$ back:

$\mathrm{cosec}\theta =2⟹\sin \theta =\frac{1}{2}$

$\mathrm{cosec}\theta =-\frac{2}{\sqrt{3}}⟹\sin \theta =-\frac{\sqrt{3}}{2}$

The solutions are: $-\frac{7\pi}{6}, -\frac{2\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{4\pi}{3}$.

Therefore, the number of solutions is 6.

$\sqrt{3}{\mathrm{cosec}}^2\theta -2\left(\sqrt{3}-1\right)\mathrm{cosec}\theta -4=0$

Let $x = \csc \theta$.

$\sqrt{3} x^2 - 2(\sqrt{3} - 1) x - 4 = 0$

$\sqrt{3}x(x - 2) + 2(x - 2) = 0$

$(\sqrt{3}x + 2)(x - 2) = 0$

So, the solutions for $x$ are: \(x=2 \) or \(x=-\frac{2}{\sqrt{3}}\)

Substituting $x=\mathrm{cosec}\theta$ back:

$\mathrm{cosec}\theta =2⟹\sin \theta =\frac{1}{2}$

$\mathrm{cosec}\theta =-\frac{2}{\sqrt{3}}⟹\sin \theta =-\frac{\sqrt{3}}{2}$

The solutions are: $-\frac{7\pi}{6}, -\frac{2\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{4\pi}{3}$.

Therefore, the number of solutions is 6.

Given, \(\sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi]\)

\(\Rightarrow \sqrt{3}\left(2 \cos ^2 \theta-1\right)+8 \cos \theta+3 \sqrt{3}=0\)

\(\Rightarrow 2 \sqrt{3} \cos ^2 \theta+8 \cos \theta+2 \sqrt{3}=0\)

\(\Rightarrow (\sqrt{3} \cos +1)(\cos \theta+\sqrt{3})=0\)

\(\Rightarrow \cos \theta=-\frac{1}{\sqrt{3}}\) and \( \cos \theta=-\sqrt{3}\) (rejected)

\(\therefore \theta\) will have \(5\) values in \([-3 \pi, 2 \pi]\)

Total number of solutions are \(5\).

Given:

\( f(\theta)=4\left(\sin ^4\left(\frac{7 \pi}{2}-\theta\right)+\sin ^4(11 \pi+\theta)\right) -2\left(\sin ^6\left(\frac{3 \pi}{2}-\theta\right)+\sin ^6(9 \pi-\theta)\right)\)

\(\Rightarrow f(\theta)=4\left(\cos ^4(\theta)+\sin ^4(\theta)\right)-2\left(\cos ^6 \theta+\sin ^6 \theta\right)\)

\(\Rightarrow f(\theta)=4\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)-2\left(1-3 \sin ^2 \theta \cos ^2 \theta\right)\)

\(\Rightarrow f(\theta)=2-2 \sin ^2 \theta \cos ^2 \theta\)

\(\Rightarrow f(\theta)=2-\frac{\sin ^2(2 \theta)}{2}\)

So, \(\alpha=f(\theta)_{\max }=2\),

\(\beta=f(\theta)_{\min }=\frac{3}{2}\)

\(\Rightarrow \alpha+2 \beta=5\)

Let \(y=\sin x(\sin x+\cos x)=\sin ^2 x+\sin x \cos x\)

\(y=\frac{1-\cos 2 x}{2}+\frac{\sin 2 x}{2}=\frac{1}{2}+\frac{\sin 2 x-\cos 2 x}{2}\)

Since \(\sin 2 x-\cos 2 x \in[-\sqrt{2}, \sqrt{2}]\), we get:

\(y \in\left[\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]\)

Therefore, possible integer values of \(y\) are \(0\) and \(1\)

For \(y=0, \sin x(\sin x+\cos x)=0\)

\(\Rightarrow \sin x=0\) or \(\sin x+\cos x=0\)

\(\sin x=0 \Rightarrow x=-\pi, 0, \pi\)

\(\sin x+\cos x=0 \Rightarrow \tan x=-1 \Rightarrow x=-\frac{\pi}{4}, \frac{3 \pi}{4}\)

So, for \(y=0\), number of solutions \(=5\)

For \(y=1, \sin ^2 x+\sin x \cos x=1\)

\(\Rightarrow 1-\cos 2 x+\sin 2 x=2 \Rightarrow \sin 2 x-\cos 2 x=1\)

\(\sqrt{2} \sin \left(2 x-\frac{\pi}{4}\right)=1 \Rightarrow \sin \left(2 x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\)

So, \(2 x-\frac{\pi}{4}=\frac{\pi}{4}+2 k \pi\) or \(2 x-\frac{\pi}{4}=\frac{3 \pi}{4}+2 k \pi\)

Thus, \(x=\frac{\pi}{4}+k \pi\) or \(x=\frac{\pi}{2}+k \pi\).

In \( [-\pi, \pi], x=-\frac{3 \pi}{4}, \frac{\pi}{4},-\frac{\pi}{2}, \frac{\pi}{2}\)

So, for \(y=1\), number of solutions \(=4\)

Hence, \(n(S)=5+4=9\)

\(K=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right)\)

We know \(\sin x \sin \left(60^{\circ}-x\right) \sin \left(60^{\circ}+x\right)=\frac{\sin 3 x}{4}\)

Here \(x=\frac{\pi}{18}=10^{\circ}\)

Hence we get \(K=\frac{\sin \frac{\pi}{6}}{4}=\frac{1 / 2}{4}=\frac{1}{8}\)

Therefore $\sin \left(\frac{10K\pi }{3}\right)$\(=\sin \left(10 \times \frac{1}{8} \cdot \frac{\pi}{3}\right)=\sin \frac{5 \pi}{12}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

(D)

Let \(E=\frac{\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}}\)

\(\because \cos \theta \cos \left(60^{\circ}-\theta\right) \cos \left(60^{\circ}+\theta\right)=\frac{1}{4} \cos 3 \theta\)

Here \(\theta=20^{\circ}\)

$\therefore E=\frac{\frac{\sqrt{3}}{\sin {20}^{°}}-\frac{1}{\cos {20}^{°}}}{\frac{1}{2}\cdot \frac{1}{4}\cdot \cos {60}^{°}}$

$=\frac{\left(\sqrt{3}\cos {20}^{°}-\sin {20}^{°}\right)}{\cos {20}^{°}\cdot \sin {20}^{°}}\times 16$

$=\frac{\left(\frac{\sqrt{3}}{2}\cos {20}^{°}-\frac{1}{2}\sin {20}^{°}\right)}{2\cos {20}^{°}\cdot \sin {20}^{°}}\times 32\times 2$

$=\frac{\sin 60°\cos 20°-\cos 60°\sin 20°}{\sin 40°}\times 64$

$=\frac{\sin {40}^{°}}{\sin {40}^{°}}\times 64=64$

\(4\cos\alpha+5\sin\alpha=1\)

\(-\frac{\pi}{2}<\alpha<\frac{\pi}{2}\),

\(\cos\alpha>0\).

\(t=\tan\alpha\).

\(\sin\alpha=t\cos\alpha\)

Substituting, \((4+5t)\cos\alpha=1\)

\(\cos\alpha=\frac{1}{4+5t}\)

\(\sec^2\alpha=1+\tan^2\alpha\),

\((4+5t)^2=1+t^2\)

\(16+40t+25t^2=1+t^2\)

\(24t^2+40t+15=0\)

\(t=\frac{-40\pm\sqrt{1600-1440}}{48}\)

\(=\frac{-40\pm4\sqrt{10}}{48}\)

\(=\frac{-10\pm\sqrt{10}}{12}\)

\(\cos\alpha=\frac{1}{4+5t}>0\)

\(t=\frac{-10-\sqrt{10}}{12}\),

\(4+5t<0\),

which gives \(\cos\alpha<0\), not possible.

Therefore \(t=\frac{-10+\sqrt{10}}{12}\)

\(=\frac{\sqrt{10}-10}{12}\)

Consider \(\tan (A+B)\)

\(= \frac{\tan A+\tan B}{1-\tan A \tan B}\)

\(=\frac{\frac{1}{\sqrt{x\left(x^2+x+1\right)}}+\frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1-\frac{1}{x^2+x+1}} \)
\( \Rightarrow \tan (\mathrm{A}+\mathrm{B})=\frac{(1+x)\left(\sqrt{x^2+x+1}\right)}{\left(x^2+x\right)(\sqrt{x})}\)
\(\Rightarrow \tan (A+B)=\frac{\sqrt{x^2+x+1}}{x \sqrt{x}}=\tan C \)
\( A+B=C\)

\(\tan^2\theta=\sec^2\theta-1\),

\(2(\sec^2\theta-1)-5\sec\theta=1\)

\(2\sec^2\theta-5\sec\theta-3=0\)

\(x=\sec\theta\).

\(2x^2-5x-3=0\)

\((2x+1)(x-3)=0\)

\(\sec\theta=3\)

\(\sec\theta=-\frac12\)

Since \(\sec\theta=-\frac12\) is impossible,

\(\sec\theta=3\)

\(\cos\theta=\frac13\)

\(\alpha=\cos^{-1}\left(\frac13\right)\)

where \(0<\alpha<\frac{\pi}{2}\).

The solutions are \(\theta=2m\pi\pm\alpha\)

For every interval of length \(2\pi\), there are exactly \(2\) solutions.

Also in \(\left[0,\frac{5\pi}{2}\right]\): \(3\) solutions.

In \(\left[0,\frac{9\pi}{2}\right]\): \(5\) solutions.

In \(\left[0,\frac{13\pi}{2}\right]\): \(7\) solutions.

Therefore the least value of \(n\) is \(n=13\).

\(\sum_{k=1}^{n}\frac{k}{2^k}
=\frac{\frac12-(n+1)\left(\frac12\right)^{n+1}+n\left(\frac12\right)^{n+2}}{\left(1-\frac12\right)^2}\)

For \(n=13\),

\(\sum_{k=1}^{13}\frac{k}{2^k}
=2-\frac{15}{2^{13}}\)

\(=\frac{1}{2^{13}}(2^{14}-15)\)

\(\tan^2\theta=\sec^2\theta-1\),

\(2(\sec^2\theta-1)-5\sec\theta=1\)

\(2\sec^2\theta-5\sec\theta-3=0\)

\(x=\sec\theta\).

\(2x^2-5x-3=0\)

\((2x+1)(x-3)=0\)

\(\sec\theta=3\)

\(\sec\theta=-\frac12\)

Since \(\sec\theta=-\frac12\) is impossible,

\(\sec\theta=3\)

\(\cos\theta=\frac13\)

\(\alpha=\cos^{-1}\left(\frac13\right)\)

where \(0<\alpha<\frac{\pi}{2}\).

The solutions are \(\theta=2m\pi\pm\alpha\)

For every interval of length \(2\pi\), there are exactly \(2\) solutions.

Also in \(\left[0,\frac{5\pi}{2}\right]\): \(3\) solutions.

In \(\left[0,\frac{9\pi}{2}\right]\): \(5\) solutions.

In \(\left[0,\frac{13\pi}{2}\right]\): \(7\) solutions.

Therefore the least value of \(n\) is \(n=13\).

\(\sum_{k=1}^{n}\frac{k}{2^k}
=\frac{\frac12-(n+1)\left(\frac12\right)^{n+1}+n\left(\frac12\right)^{n+2}}{\left(1-\frac12\right)^2}\)

For \(n=13\),

\(\sum_{k=1}^{13}\frac{k}{2^k}
=2-\frac{15}{2^{13}}\)

\(=\frac{1}{2^{13}}(2^{14}-15)\)

1. Simplify the trigonometric expression

Let the given expression be $E$.

$E = \left(\sin \frac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos \frac{11x}{2}\right)(\sin 6x + \cos 6x)$

Expand the terms:

$E = \sin \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x + \cos \frac{11x}{2} \sin 6x + \cos \frac{11x}{2} \cos 6x$

Rearrange the terms to group familiar identities:

$E = \left(\cos \frac{11x}{2} \cos 6x + \sin \frac{11x}{2} \sin 6x\right) + \left(\cos \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x\right)$

We recognize the cosine difference formula

$\cos(A-B) = \cos A \cos B + \sin A \sin B$ and

the sine difference formula

$\sin(B-A) = \sin B \cos A - \cos B \sin A$.

Let $A = \frac{11x}{2}$ and $B = 6x$.

The first part is $\cos\left(\frac{11x}{2} - 6x\right)$.

The second part is $\sin\left(6x - \frac{11x}{2}\right)$.

Calculate the arguments:

$A - B = \frac{11x}{2} - \frac{12x}{2} = -\frac{x}{2}$

$B - A = \frac{12x}{2} - \frac{11x}{2} = \frac{x}{2}$

Substitute these back into the expression:

$E = \cos\left(-\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)$

Since cosine is an even function

, $\cos(-\theta) = \cos(\theta)$:

$E = \cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)$

1. Determine $\sin x$ and $\cos x$ from the given $\cot x$

We are given $\cot x = -\frac{5}{\sqrt{11}}$ and $\frac{\pi}{2} < x < \pi$.

The condition $\frac{\pi}{2} < x < \pi$ means $x$ is

in the second quadrant. In the second quadrant,

$\sin x > 0$ and $\cos x < 0$.

We can construct a right triangle where

the adjacent side is 5 and the opposite

side is $\sqrt{11}$ (ignoring the sign for now).

The hypotenuse $h$ is calculated

using the Pythagorean theorem:

$h = \sqrt{5^2 + (\sqrt{11})^2} = \sqrt{25 + 11} = \sqrt{36} = 6$

Now, we can find $\sin x$ and $\cos x$:

$\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{11}}{6}$

$\cos x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{6}$

Applying the signs for the second quadrant:

$\sin x = \frac{\sqrt{11}}{6}$

$\cos x = -\frac{5}{6}$

1. Determine $\sin\left(\frac{x}{2}\right)$ and $\cos\left(\frac{x}{2}\right)$

Given $\frac{\pi}{2} < x < \pi$, we can find the range for $\frac{x}{2}$:

$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$

This means $\frac{x}{2}$ is in the first quadrant.

In the first quadrant, both $\sin\left(\frac{x}{2}\right)$ and

$\cos\left(\frac{x}{2}\right)$ are positive.

We use the half-angle formulas:

$\sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos x}{2}$

$\cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos x}{2}$

Substitute the value of $\cos x = -\frac{5}{6}$:

$\sin^2\left(\frac{x}{2}\right) = \frac{1 - \left(-\frac{5}{6}\right)}{2} = \frac{1 + \frac{5}{6}}{2} = \frac{\frac{6+5}{6}}{2} = \frac{\frac{11}{6}}{2} = \frac{11}{12}$

Since $\sin\left(\frac{x}{2}\right) > 0$:

$\sin\left(\frac{x}{2}\right) = \sqrt{\frac{11}{12}} = \frac{\sqrt{11}}{\sqrt{4 \cdot 3}} = \frac{\sqrt{11}}{2\sqrt{3}}$

And for $\cos\left(\frac{x}{2}\right)$:

$\cos^2\left(\frac{x}{2}\right) = \frac{1 + \left(-\frac{5}{6}\right)}{2} = \frac{1 - \frac{5}{6}}{2} = \frac{\frac{6-5}{6}}{2} = \frac{\frac{1}{6}}{2} = \frac{1}{12}$

Since $\cos\left(\frac{x}{2}\right) > 0$:

$\cos\left(\frac{x}{2}\right) = \sqrt{\frac{1}{12}} = \frac{1}{\sqrt{4 \cdot 3}} = \frac{1}{2\sqrt{3}}$

1. Calculate the final expression

Now substitute the values of $\sin\left(\frac{x}{2}\right)$ and $\cos\left(\frac{x}{2}\right)$

into the simplified expression $E$:

$E = \cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)$

$E = \frac{1}{2\sqrt{3}} + \frac{\sqrt{11}}{2\sqrt{3}}$

$E = \frac{1 + \sqrt{11}}{2\sqrt{3}}$

Comparing this result with the given options:

A: $\frac{\sqrt{11}-1}{2\sqrt{3}}$

B: $\frac{\sqrt{11}+1}{2\sqrt{3}}$

C: $\frac{\sqrt{11}+1}{3\sqrt{2}}$

D: $\frac{\sqrt{11}-1}{3\sqrt{2}}$

The calculated value matches option B.

The final answer is $\boxed{\text{B } \frac{\sqrt{11}+1}{2\sqrt{3}}}$.

\(p=12 \cos x-3 \cos ^2 x \)
\(p=-3\left(\cos ^2 x-4 \cos x\right) \)
\(p=-3\left((\cos x-2)^2-4\right)\)
Put \(\cos x=-1 \Rightarrow p=-15\)
Put \(\cos x=1 \)

\(\Rightarrow p=9\)

\(-15 \leq p \leq 9\)

So integral values of \(p\) are \(-15,-14, \ldots, 9\)

Sum \(=\frac{25}{2}(-15+9) = -75\)

\( f(x)=3 \sqrt{x-2}+\sqrt{4-x} \)
\( x-2 \geq 0 \quad \& \quad 4-x \geq 0 \)
\( \therefore x \in[2,4] \)
Let \( x=2 \sin ^2 \theta+4 \cos ^2 \theta \)
\( \therefore f(x)=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \)
\( \therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2}\)

\( \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20} \)

\( \therefore \alpha=\sqrt{2}, \beta=\sqrt{20} \)
\( \alpha^2+2 \beta^2=2+40=42\)

\(\begin{equation}
\begin{aligned}
& 2 \sin ^2 \theta=\cos 2 \theta \\
& 2 \cos ^2 \theta=3 \sin \theta \\
& \Rightarrow \text { Adding, } \\
& 2= 1-2 \sin ^2 \theta+3 \sin \theta \\
& \Rightarrow 2 \sin ^2 \theta-3 \sin \theta+1=0 \\
& 2 \sin ^2 \theta-2 \sin \theta-\sin \theta+1=0 \\
& 2 \sin \theta(\sin \theta-1)-1(\sin \theta-1)=0 \\
& \sin \theta=1, \frac{1}{2} \text { but } 2 \sin ^2 \theta=\cos 2 \theta=2 \\
& \text { but not is not possible } \\
& \Rightarrow \theta=\frac{\pi}{6},\left(\pi-\frac{\pi}{6}\right) \\
& \Rightarrow \text { Sum of all values }=\frac{\pi}{6}+\pi-\frac{\pi}{6}=\pi
\end{aligned}
\end{equation}\)

\( x^2-2 x-5=0\)

\( \tan \mathrm{~A}+\tan \mathrm{B}=2\)

\( \tan \mathrm{A} \tan \mathrm{B}=-5\)

\( \therefore \tan (\mathrm{~A}+\mathrm{B})=\frac{2}{1-(-5)}=\frac{1}{3}\)

\( \Rightarrow \cos (\mathrm{~A}+\mathrm{B})=\frac{3}{\sqrt{10}}\)

\( \therefore 20\left(\sin ^2\left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right)\right)=\frac{20}{2}(1-\cos (\mathrm{A}+\mathrm{B}))\)

\( =10\left(1-\frac{3}{\sqrt{10}}\right)=10-3 \sqrt{10}\)

Given, \(\frac{\tan (A-B)}{\tan A}+\frac{\sin ^2 C}{\sin ^2 A}=1\)

$\Rightarrow \frac{\tan A-\tan B}{\left(1+\tan A\tan B\right)\tan A}+\frac{1+{\cot }^2 A}{1+{\cot }^{2}C}=1$

Put $\tan A=x,\tan B=y,\tan C=z$

$\therefore \frac{x-y}{\left(1+xy\right)x}+\frac{\left(x^2+1\right)z^2}{x^2\left(z^2+1\right)}=1$

$\therefore x\left(x-y\right)\left(z^2+1\right)+z^2\left(1+x^2\right)\left(1+xy\right)=\left(1+xy\right)x^2\left(1+z^2\right)$

After solving we get

$z^2=xy$ $\left[\because 1+x^2\ne 0\right]$

$\therefore {\tan }^{2}C=\tan A\cdot \tan B$

Hence, $\tan A,\tan C,\tan B$ are in G.P.

$\frac{3\cos 2x+{\cos }^{3}2x}{{\cos }^{6}x-{\sin }^{6}x}=x^3-x^2+6$

$\Rightarrow \frac{\cos 2x(3+{\cos }^{2}2x)}{\cos 2x(1-{\sin }^{2}x{\cos }^{2}x)}=x^3-x^2+6$

$\Rightarrow \frac{4(3+{\cos }^{2}2x)}{4-{\sin }^{2}2x}=x^3-x^2+6$

$\Rightarrow \frac{4(3+{\cos }^{2}2x)}{3+{\cos }^{2}2x}=x^3-x^2+6$

$\Rightarrow x^3-x^2+2=0$

$\Rightarrow (x+1)(x^2-2x+2)=0$

Since roots of $x^2-2x+2=0$ are not real

Therefore sum of real solutions $=-1$

$\frac{3\cos 2x+{\cos }^{3}2x}{{\cos }^{6}x-{\sin }^{6}x}=x^3-x^2+6$

$\Rightarrow \frac{\cos 2x(3+{\cos }^{2}2x)}{\cos 2x(1-{\sin }^{2}x{\cos }^{2}x)}=x^3-x^2+6$

$\Rightarrow \frac{4(3+{\cos }^{2}2x)}{4-{\sin }^{2}2x}=x^3-x^2+6$

$\Rightarrow \frac{4(3+{\cos }^{2}2x)}{3+{\cos }^{2}2x}=x^3-x^2+6$

$\Rightarrow x^3-x^2+2=0$

$\Rightarrow (x+1)(x^2-2x+2)=0$

Since roots of $x^2-2x+2=0$ are not real

Therefore sum of real solutions $=-1$

\( \cos 2 \theta \cos \frac{\theta}{2}+\cos \frac{5 \theta}{2}=2 \cos ^3 \frac{5 \theta}{2} \)
\( \frac{1}{2}\left(2 \cos 2 \theta \cos \frac{\theta}{2}\right)+\cos \frac{5 \theta}{2}=2 \cos ^3 \frac{5 \theta}{2} \)
\( \frac{1}{2}\left(\cos \frac{5 \theta}{2}+ \cos \frac{3 \theta}{2}\right)=2 \cos ^3 \frac{5 \theta}{2}\)

\(\frac{3}{2} \cos \frac{5 \theta}{2}+\frac{1}{2} \cos \frac{3 \theta}{2}=2 \cos ^3 \frac{5 \theta}{2}\)

\(\frac{3}{2} \cos \frac{5 \theta}{2}+\frac{1}{2} \cos \frac{3 \theta}{2}=\frac{3}{2} \cos \frac{5 \theta}{2}+\frac{1}{2} \cos \frac{15 \theta}{2}\)

\( \cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2} \)

\( \cos \frac{15 \theta}{2}-\cos \frac{3 \theta}{2}=0 \)
\( 2 \sin 3\theta \sin \frac{9 \theta}{2}=0 \)
\( 3 \theta=\mathrm{n} \pi \text { or } \frac{9 \theta}{2}=\mathrm{m} \pi\)
\(\theta=\frac{\mathrm{n} \pi}{3} \text { or } \theta=\frac{2 \mathrm{~m} \pi}{9} \)
\( \theta=\left\{-\frac{\pi}{2}, \frac{\pi}{3}, 0\right\} \)
\( \theta=\left\{-\frac{4 \pi}{9}, \frac{-2 \pi}{9}, \frac{4 \pi}{9}, \frac{2 \pi}{9}\right\}\)

Let \(f(\theta)=\cos ^2 \theta-6 \sin \theta \cos \theta+3 \sin ^2 \theta+2\)

$\Rightarrow f\left(\theta \right)=\frac{1+\cos 2\theta }{2}-3\sin 2\theta +3\left(\frac{1-\cos 2\theta }{2}\right)+2$

$\mathit{\Rightarrow }f\left(\theta \right)=4-3\sin 2\theta -\cos 2\theta$

Hence, $f\left(\theta \right)\in [4-\sqrt{10},4+\sqrt{10}]$

Given, \(\frac{\tan (A-B)}{\tan A}+\frac{\sin ^2 C}{\sin ^2 A}=1\)

$\Rightarrow \frac{\tan A-\tan B}{\left(1+\tan A\tan B\right)\tan A}+\frac{1+{\cot }^2 A}{1+{\cot }^{2}C}=1$

Put $\tan A=x,\tan B=y,\tan C=z$

$\therefore \frac{x-y}{\left(1+xy\right)x}+\frac{\left(x^2+1\right)z^2}{x^2\left(z^2+1\right)}=1$

$\therefore x\left(x-y\right)\left(z^2+1\right)+z^2\left(1+x^2\right)\left(1+xy\right)=\left(1+xy\right)x^2\left(1+z^2\right)$

After solving we get

$z^2=xy$ $\left[\because 1+x^2\ne 0\right]$

$\therefore {\tan }^{2}C=\tan A\cdot \tan B$

Hence, $\tan A,\tan C,\tan B$ are in G.P.

\(\left(\sin 70^{\circ}\right)\left(\cot 10^{\circ} \cot 70^{\circ}-1\right)=\left(\sin 70^{\circ}\right)\left(\frac{\cos 10^{\circ} \cos 70^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}}-1\right)\)

\(=\left(\sin 70^{\circ}\right)\left(\frac{\cos 10^{\circ} \cos 70^{\circ}-\sin 10^{\circ} \sin 70^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}}\right)\)

\(=\left(\sin 70^{\circ}\right)\left(\frac{\cos 80^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}}\right)\)

\(=\frac{\sin 10^{\circ} \cdot \sin 70^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}}=1\)

\(2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0\)
\( 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \)
\( 6 \sin x-4=0 \)
\( \sin x=\frac{2}{3}\)
\( n=5 \) (in the given interval)
\( x^2+5 x+2=0 \)
\( x=\frac{-5 \pm \sqrt{17}}{2}\)

Required interval \((-\infty, 0)\)

\(2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0\)
\( 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \)
\( 6 \sin x-4=0 \)
\( \sin x=\frac{2}{3}\)
\( n=5 \) (in the given interval)
\( x^2+5 x+2=0 \)
\( x=\frac{-5 \pm \sqrt{17}}{2}\)

Required interval \((-\infty, 0)\)

Let \(f(\theta)=\cos ^2 \theta-6 \sin \theta \cos \theta+3 \sin ^2 \theta+2\)

$\Rightarrow f\left(\theta \right)=\frac{1+\cos 2\theta }{2}-3\sin 2\theta +3\left(\frac{1-\cos 2\theta }{2}\right)+2$

$\mathit{\Rightarrow }f\left(\theta \right)=4-3\sin 2\theta -\cos 2\theta$

Hence, $f\left(\theta \right)\in [4-\sqrt{10},4+\sqrt{10}]$

\( (4-\sqrt{3}) \sin x-2 \sqrt{3} \cos ^2 x=\frac{-4}{1+\sqrt{3}}, x \in\left[-2 \pi, \frac{5 \pi}{2}\right] \)
\( \Rightarrow(4-\sqrt{3}) \sin x-2 \sqrt{3}\left(1-\sin ^2 x\right)=2(1-\sqrt{3})\)
\(\Rightarrow 2 \sqrt{3} \sin ^2 \mathrm{x}+4 \sin \mathrm{x}-\sqrt{3} \sin \mathrm{x}-2=0 \)
\( \Rightarrow(2 \sin x-1)(\sqrt{3} \sin x+2)=0 \)
\( \Rightarrow \sin \mathrm{x}=\frac{1}{2} \)
\( \therefore \) Number of solution \(=5\)

Given: $\cot x=\frac{5}{12}\Rightarrow \cos x=\frac{-5}{13}$

$\Rightarrow \frac{-5}{13}=2{\cos }^2\frac{x}{2}-1$

\(\Rightarrow \cos \left(\frac{x}{2}\right)=-\frac{2}{\sqrt{13}}\) or \(\frac{2}{\sqrt{13}}\) (rejected)

$\left\{\because \frac{x}{2}\in \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\right\}$

\(\sin 7 x\left(\cos \frac{13 x}{2}+\sin \frac{13 x}{2}\right)+\cos 7 x\left(\cos \frac{13 x}{2}-\sin \frac{13 x}{2}\right)\)

\(=\left(\sin 7 x \frac{\sin 13 x}{2}+\cos 7 x \frac{\cos 13 x}{2}\right)+\left(\sin 7 x \frac{\cos 13 x}{2}-\cos 7 x \frac{\sin 13 x}{2}\right)\)

\(=\cos \left(7 x-\frac{13 x}{2}\right)+\sin \left(7 x-\frac{13 x}{2}\right)\)

\(=\cos \frac{x}{2}+\sin \left(\frac{x}{2}\right)\)

\(=\frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}}=\frac{1}{\sqrt{13}}\)

\(
\begin{aligned}
& \mathrm{f}(\mathrm{x})=6+\frac{16}{4} \cos 3 \mathrm{x} \cdot \sin 3 \mathrm{x} \cdot \cos 6 \mathrm{x} \\
& =6+2 \sin 6 \mathrm{x} \cdot \cos 6 \mathrm{x} \\
& \mathrm{f}(\mathrm{x}) =6+\sin 12 \mathrm{x} \\
& \mathrm{f}(\mathrm{x}) \in [5,7] \\
& 3 \mathrm{x}+4 \mathrm{y}+12=0 \\
& d=\left|\frac{15+28+12}{5}\right|=\left|\frac{55}{5}\right|=11
\end{aligned}
\)

$10{\sin }^4\theta +15{\cos }^4\theta =6$

\(\Rightarrow 10\left[1-2 \sin ^2 \theta \cos ^2 \theta\right]+5 \cos ^4 \theta=6\)

\(\Rightarrow 25 \cos ^4 \theta-20 \cos ^2 \theta+4=0\)

\(\Rightarrow\left(5 \cos ^2 \theta-2\right)^2=0\)

$\Rightarrow {\cos }^2\theta =\frac{2}{5}$ and ${\sin }^2\theta =\frac{3}{5}$

$\Rightarrow \frac{27{cosec}^6\theta +8{\sec }^6\theta }{16{\sec }^8\theta }=\frac{27·{\left(\frac{5}{3}\right)}^3+8·{\left(\frac{5}{2}\right)}^3}{16·{\left(\frac{5}{4}\right)}^4}=\frac{2}{5}$

\[
\begin{aligned}
& 4\left(\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\right)
-3\left(\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\right)=1 \\
& \text{Let } \tan\frac{\theta}{2}=t. \\
& \frac{4-4t^2-6t}{1+t^2}=1 \\
& 4-4t^2-6t=1+t^2 \\
& \Rightarrow 5t^2+6t-3=0 \\
& \Rightarrow t=\frac{-6\pm\sqrt{36-4(5)(-3)}}{10} \\
& =\frac{-6\pm\sqrt{96}}{10} \\
& =\frac{-6\pm4\sqrt{6}}{10} \\
& t=\frac{-3+2\sqrt{6}}{5} \\
& \cos\theta=\frac{1-t^2}{1+t^2} \\
& =\frac{1-\left(\frac{2\sqrt{6}-3}{5}\right)^2}
{1+\left(\frac{2\sqrt{6}-3}{5}\right)^2} \\
& =\frac{1-\left(\frac{24+9-12\sqrt{6}}{25}\right)}
{1+\left(\frac{24+9-12\sqrt{6}}{25}\right)} \\
& =\frac{25-33+12\sqrt{6}}{25+33-12\sqrt{6}} \\
& =\frac{12\sqrt{6}-8}{58-12\sqrt{6}} \\
& =\frac{6\sqrt{6}-4}{29-6\sqrt{6}}
\cdot
\frac{29+6\sqrt{6}}{29+6\sqrt{6}} \\
& =\frac{100+150\sqrt{6}}{625} \\
& =\frac{4+6\sqrt{6}}{25}
\cdot
\frac{4-6\sqrt{6}}{4-6\sqrt{6}} \\
& =\frac{-200}{25(4-6\sqrt{6})} \\
& =\frac{-8}{4-6\sqrt{6}} \\
& =\frac{4}{3\sqrt{6}-2}
\end{aligned}
\]

We know that

$\cos \left(\theta \right)cos\left(60°-\theta \right)\cos \left(60°+\theta \right)=\frac{1}{4}\cos 3\theta$

\(|\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8}\)

$\Rightarrow \left|\frac{1}{4}\cos 3\theta \right|\le \frac{1}{8}$

\( \Rightarrow|\cos 3 \theta| \leq \frac{1}{2} \)
\( \Rightarrow-\frac{1}{2} \leq \cos 3 \theta \leq \frac{1}{2}\)

⇒ maximum value of \(\cos 3 \theta=\frac{1}{2}\), here

\( \Rightarrow 3 \theta=2 \mathrm{n} \pi \pm \frac{\pi}{3} \)
\( \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9}\)

As \(\theta \in[0,2 \pi]\) possible values are

\(\theta=\left\{\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9}\right\}\)
Whose sum is \(\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}+\frac{11 \pi}{9}+\frac{13 \pi}{9}+\frac{17 \pi}{9}=\frac{54 \pi}{9}=6 \pi\)

$2\sqrt{2}{\cos }^2\theta +2\cos \theta -\sqrt{6}\cos \theta -\sqrt{3}=0$

$(2\cos \theta -\sqrt{3})(\sqrt{2}\cos \theta +1)=0$

$\cos \theta =\frac{\sqrt{3}}{2},\frac{-1}{\sqrt{2}}$

Number of solution = 8

$f\left(\theta \right)={\sin }^4\theta +{\cos }^2\theta$

$={\sin }^2\theta \left(1-{\cos }^2\theta \right)+{\cos }^2\theta$

$={\sin }^2\theta +{\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta$

$f\left(\theta \right)=1-\frac{1}{4}{\sin }^{2}2\theta$

$\Rightarrow 0\le {\sin }^{2}2\theta \le 1$

$f\left(\theta \right)\max =1$

$f\left(\theta \right)\min =1-\frac{1}{4}=3/4$

$\therefore \text{ Range is }\left[\frac{3}{4},1\right]$

\( P=\left\{\theta \in[0,4 \pi]: \tan ^2 \theta \neq 1\right\} \)
\( S=\left\{a \in z: 2\left(\cos ^8 \theta-\sin ^8 \theta\right) \sec 2 \theta=a^2, \theta \in P\right\} \)
\( a^2=2 \cos 2 \theta\left(1-\frac{1}{2} \sin ^2 2 \theta\right) \sec 2 \theta \)
\( a^2=2-\sin ^2 2 \theta \)

Since \(\tan^2\theta\ne1\), we have \(\sin^22\theta\ne1\).
\( \therefore \) Range of \( a^2\) is \((1,2]\).
No integer value of \(a \).
\(\therefore n(S)=0\)

\(
\begin{aligned}
& \mathrm{f}(\mathrm{x})=6+\frac{16}{4} \cos 3 \mathrm{x} \cdot \sin 3 \mathrm{x} \cdot \cos 6 \mathrm{x} \\
& =6+2 \sin 6 \mathrm{x} \cdot \cos 6 \mathrm{x} \\
& \mathrm{f}(\mathrm{x}) =6+\sin 12 \mathrm{x} \\
& \mathrm{f}(\mathrm{x}) \in [5,7] \\
& 3 \mathrm{x}+4 \mathrm{y}+12=0 \\
& d=\left|\frac{15+28+12}{5}\right|=\left|\frac{55}{5}\right|=11
\end{aligned}
\)

Given : \(2 \sin ^2 \theta=\cos 2 \theta\)

\(\Rightarrow 2 \sin ^2 \theta=1-2 \sin ^2 \theta\)

\(\Rightarrow 4 \sin ^2 \theta=1\)

\(\Rightarrow \sin \theta= \pm \frac{1}{2}\)

and \(2 \cos ^2 \theta=3 \sin \theta\)

\(\Rightarrow 2 \sin ^2 \theta+3 \sin \theta-2=0\)

\(\Rightarrow(2 \sin \theta-1)(\sin \theta+2)=0\)

\(\Rightarrow \sin \theta=\frac{1}{2}\) but \(\sin \theta \neq-2\)

common equation is \(\sin \theta=\frac{1}{2}\)

\(\Rightarrow \theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])\)

Sum\(=\frac{\pi}{6}+\frac{5 \pi}{6}=\pi\)

Given : \(2 \sin ^2 \theta=\cos 2 \theta\)

\(\Rightarrow 2 \sin ^2 \theta=1-2 \sin ^2 \theta\)

\(\Rightarrow 4 \sin ^2 \theta=1\)

\(\Rightarrow \sin \theta= \pm \frac{1}{2}\)

and \(2 \cos ^2 \theta=3 \sin \theta\)

\(\Rightarrow 2 \sin ^2 \theta+3 \sin \theta-2=0\)

\(\Rightarrow(2 \sin \theta-1)(\sin \theta+2)=0\)

\(\Rightarrow \sin \theta=\frac{1}{2}\) but \(\sin \theta \neq-2\)

common equation is \(\sin \theta=\frac{1}{2}\)

\(\Rightarrow \theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])\)

Sum\(=\frac{\pi}{6}+\frac{5 \pi}{6}=\pi\)

$$\begin{gathered} \text{ Since, }\tan 30°\text{ and }\tan 15°\text{ are the} \\ \text{roots of equation }{\mathrm{x}}^2+px+\mathrm{q}=0\text{. } \\ \tan 30°+\tan 15°=-p .........\left(i\right) \\ \text{ and } \tan 30°\tan 15°=q ......\left(ii\right) \\ \text{We know that,} \\ \tan {45}^o=\tan \left({30}^o+{15}^0\right) \\ 1=\frac{\tan 30°+\tan 15°}{1-\tan 30°\tan 15°} \\ 1=\frac{-p}{1-q} \\ 1-q=-p \\ q-p=1 \\ On squaring, we get \\ {q}^2+p^2-2pq=1 \\ 2pq=q^2+p^2-1 \\ 2pq={\left( \tan 30°\tan 15°\right)}^2+{\left(\tan 30°+\tan 15°\right)}^2-1 \\ 2pq={\left(\frac{2-\sqrt{3}}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{3}}+2-\sqrt{3}\right)}^2-1 \\ pq=\frac{(2\sqrt{3}-3+1{)}^2+(2-\sqrt{3}{)}^2-3}{6} \\ pq=\frac{20-12\sqrt{3}}{6}=\frac{10-6\sqrt{3}}{3} \end{gathered}$$