Given trigonometric equation is
 \(\sqrt { 3 } \) sec 2x = 2
⇒ sec 2x = \(\frac { 2 }{ \sqrt { 3 } } \)
⇒ cos 2x = \(\frac { \sqrt { 3 } }{ 2 } \)
⇒ cos 2x = cos\(\left( \frac { \pi }{ 6 } \right) \)    \(\left[ \because sec2x=\frac { 1 }{ cos2x } \right] \)
⇒ \(2x=2n\pi \pm \left( \frac { \pi }{ 6 } \right) ,n\in Z\)    \(\left[ \because cos\frac { \pi }{ 6 } =\frac { \sqrt { 3 } }{ 2 } \right] \)
⇒ x = \(n\pi \pm \frac { \pi }{ 12 } ,n\in Z\)

LHS = \(\frac { cos9x-cos5x }{ sin17x-sin3x } \)
= \(\frac { -2sin\left( \frac { 9x+5x }{ 2 } \right) .sin\left( \frac { 9x-5x }{ 2 } \right) }{ -2sin\left( \frac { 17x-3x }{ 2 } \right) .cos\left( \frac { 17x-3x }{ 2 } \right) } \) 
\(\left[ \because cosC-cosD=-2sin\left( \frac { C+D }{ 2 } \right) sin\left( \frac { C-D }{ 2 } \right) sinC-sinD=-2sin\left( \frac { C-D }{ 2 } \right) cos\left( \frac { C+D }{ 2 } \right) \right] \)
= \(\frac { -sin(7x).sin(2x) }{ sin(7x).cos(10x) } =-\frac { -sin2x }{ cos10x } \) = RHS

LHS = \(cos\left( \frac { \pi }{ 13 } \right) cos\left( \frac { 9\pi }{ 13 } \right) +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 } \)
\(cos\left( \frac { 9\pi }{ 13 } +\frac { \pi }{ 13 } \right) +cos\left( \frac { 9\pi }{ 13 } -\frac { \pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) +cos\left( \frac { 5\pi }{ 13 } \right) \)    [2cos A cos B = cos(A + B) + cos(A - B)
= \(cos\left( \frac { 10\pi }{ 13 } \right) +cos\left( \frac { 8\pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) +cos\left( \frac { 5\pi }{ 13 } \right) \)
= \(cos\left( \pi -\frac { 3\pi }{ 13 } \right) +cos\left( \pi -\frac { 5\pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) =-cos\left( \frac { 3\pi }{ 13 } \right) -cos\left( \frac { 5\pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) +cos\left( \frac { 5\pi }{ 13 } \right) \)
= 0 = RHS 
\(\left[ \because cos(\pi -A)=-cosA \right] \)

Given cos A = \(\frac { 4 }{ 5 } \), cos B =\(\frac { 12 }{ 13 } \)
Since A, B both lie in the IV quadrant sin A, sin B are negative.
\(\therefore \ sinA=\sqrt { 1-cos^{ 2 }A } =-\sqrt { 1-\frac { 16 }{ 25 } } =-\sqrt { \frac { 9 }{ 25 } } =-\frac { 3 }{ 5 } \)
\(sinB=-\sqrt { 1-cos^{ 2 }B } =-\sqrt { 1-\frac { 144 }{ 169 } } =-\sqrt { \frac { 25 }{ 169 } } =-\frac { 5 }{ 13 } \)
Now, cos(A+B) = cos A cos B - sin A sin B
=\(\left( \frac { 4 }{ 5 } \right) \left( \frac { 12 }{ 13 } \right) -\left( \frac { -3 }{ 5 } \right) \left( \frac { -5 }{ 13 } \right) =\frac { 48 }{ 65 } -\frac { 15 }{ 65 } =\frac { 33 }{ 65 } \). 

sin\(\left( \frac { -11\pi }{ 3 } \right) =-sin\frac { 11\pi }{ 3 } \)
= \(-sin\left( \frac { 11\times 180 }{ 3 } \right) \) = -sin(660⁰)
= -sin(2 \(\times\) 360⁰ - 60⁰)
= -(-sin(60⁰))   [Angle is in the IV quadrant and sine is negative
= sin 60⁰ = \(\frac { \sqrt { 3 } }{ 2 } \).

\(\frac { 2\pi }{ 5 } \)
\(\frac { 2\pi }{ 5 } \) \(\times\) \(\frac { 180 }{ \pi } \) = 2 \(\times\) 36⁰ = 72⁰

\(\frac { \pi }{ 9 } \)
\(\frac { \pi }{ 9 }\) = \(\frac { \pi }{ 9 } \) \(​​\times\) \(\frac { 180 }{ \pi } \) = 20^o

-55⁰
Since the given angle is negative, it moves in the clockwise direction
∴ -55⁰ lies in the IV quadrant

Sin 25⁰  is an acute angle, 25⁰ lies in the I quadrant

Given sec θ + tan θ = p ...(1)
We know sec²θ-tan²θ = 1
(sec θ + tan θ) (sec θ - tan θ) = 1
p(sec θ - tan θ) = 1
sec θ - tan θ = \(\frac{1}{p}\)...(2)
(1)+(2)➝ (sec θ + tan θ) + (sec θ - tan θ) = p+\(\frac{1}{p}\)
2sec θ = \(\frac{p^2+1}{2p}\)...(3)
(1)-(2)⟶ (sec θ + tan θ) - (sec θ - tan θ) = p-\(\frac{1}{p}\)
2tan θ = \(\frac{p^2-1}{2p}\)
tan θ = \(\frac{p^2-1}{2p}\)...(4)
(4)+(3) gives,
\(\frac{tan\theta}{sec\theta}=\frac{p^2-1}{2p}\div\frac{p^2+1}{2p}\)
\(\frac{sin\theta}{cos\theta.\frac{1}{cos\theta}}=\frac{p^2-1}{2p}\times\frac{2p}{p^2+1}=\frac{p^2-1}{p^2+1}\)
Sin \(\theta\) = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

Speed taken by the vehicle 1 = 60 km/hr.
Time \(={1\over 2}hr\)
∴ Distance = speed x time \(=\left(60\times{1\over 2}\right)=30km⇒PA=30km\)
Speed take by the 11^th vehicle = 80km/hr
Time \(={1\over 2}hr\)
∴ Distance = speed x time = 80 x \(1\over 2\) = 40km ⇒ PB = 40
Given ㄥAPB = 60⁰

Using cosine formula, c² - = a² + b² - 2ab cos C.
⇒ c² = 40² + 30² - 2(40)(30) cos 60°
c² = 160+900-2(40)(30)(1/2)
= 2500 - (40) (30) = 1300
\(⇒\ c=\sqrt{1300}=\sqrt{13\times100}=10\sqrt{13}km\)

Prove that sin (45^o + \(\theta\)) - sin (45^o - \(\theta\)) = \(\sqrt { 2 } \sin { \theta } \)
LHS = sin (45^o + \(\theta\)) - sin (45^o - \(\theta\))
= sin 45 cos \(\theta\) + cos 45 sin \(\theta\) - [sin 45 cos \(\theta\) - cos 45 sin \(\theta\)]
= \(\frac { 1 }{ \sqrt { 2 } } \cos { \theta } +\frac { 1 }{ \sqrt { 2 } } \sin { \theta } \frac { -1 }{ \sqrt { 2 } } \cos { \theta } +\frac { 1 }{ \sqrt { 2 } } \sin { \theta } \)
=  \(\frac { 2\sin { \theta } }{ \sqrt { 2 } } \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 2\sqrt { 2 } \sin { \theta } }{ 2 } =\sqrt { 2 } \sin { \theta } \) = RHS
Hence proved.

LHS = sin x + sin 2x + sin 3x
= (sin x + sin 3x) + sin 2x
\(=2sin\left( \frac { x+3x }{ 2 } \right) cos\left( \frac { x-3x }{ 2 } \right) +sin2x\)
= 2sin 2x.cos(-x) + sin 2x
= 2sin 2x + cos x + sin 2x
= sin 2x(1 + 2cos x) = RHS

450⁰
450⁰ = -720⁰ +270⁰
\(\Rightarrow \) -450⁰ - 270⁰ = -720⁰
∴ Coterminal angle of (-450) is 270⁰

\(LHS=\frac { (cos\theta -cos3\theta )(sin8\theta +sin2\theta ) }{ (sin5\theta -sin\theta )(cos4\theta -cos6\theta ) } \)
\(=\frac { 2sin\left( \frac { \theta +3\theta }{ 2 } \right) sin\left( \frac { 3\theta -\theta }{ 2 } \right) .2sin\left( \frac { 8\theta +2\theta }{ 2 } \right) cos\left( \frac { 8\theta -2\theta }{ 2 } \right) }{ 2cos\left( \frac { 5\theta +\theta }{ 2 } \right) sin\left( \frac { 5\theta -\theta }{ 2 } \right) .2sin\left( \frac { 4\theta +6\theta }{ 2 } \right) sin\left( \frac { 6\theta -4\theta }{ 2 } \right) } \)
\(=\frac { sin2\theta .sin\theta .sin5\theta .cos3\theta }{ cos3\theta .sin2\theta .sin5\theta .sin\theta } =1=RHS\)