We have
L.H.S. = 2cos\({\pi\over13}\)cos\({9\pi\over13}\)+cos\({3\pi\over13}\)+cos\({5\pi\over13}\)
\(=cos({9\pi\over13}+{\pi\over13})+cos({{9\pi\over13}-{\pi\over13}})+cos{3\pi\over13}+cos{5\pi\over13}\)
\([\because 2cos\ Acos \ B=cos (A+B)+cos(A-B)]\)
\(=cos{10\pi\over 13}+cos{8\pi\over13}+cos{3\pi\over13}+cos{5\pi\over13}\)
\(=cos({\pi-{3\pi\over13}})+cos({\pi-{5\pi\over13}})+cos{3\pi\over13}+cos{5\pi\over13}\)
[\(\because\)cos (\(\pi\) - \(\theta\)) = - cos \(\theta\)]
\(=-cos{3\pi\over13}-cos{5\pi\over13}+cos{3\pi\over13}+cos{5\pi\over13}=0\)
=R.H.S

sin 2x + cos x = 0
\(\Rightarrow\)2 sin x cos x + cos x = 0
\(\Rightarrow\) cos x (2 sin x + 1) = 0
\(\Rightarrow\)Either cos x = 0 or 2 sin x + 1 = 0
\(\Rightarrow\) X = (2n+ 1)\({\pi\over 2}\) or sin x=\(-{1\over2}=-sin {\pi\over6}=sin(-{\pi\over6}),n \in z\)
X = (2n+ 1)\({\pi\over 2}\) or x= n\(\pi\)+(-1)ⁿ\((-{\pi\over6})\)
X = (2n+ 1)\({\pi\over 2}\) or x=n\(\pi\)+(-1)ⁿ⁺¹\(({\pi\over6})\)
or x= n \(\pi\) +(-1)ⁿ
\(\Rightarrow x=(2n+1){\pi\over2}\) \([\because sin (\pi+{\pi\over6})=-sin{\pi\over 6}]\)
or x = n\(\pi\)+ (-1)ⁿ\({7\pi\over6}\),n \(\in\)z

We have
L.H.S. =\({sin 5x + sin 3x \over cos 5x + cos 3x }\)
\(={2sin({5x+3x\over 2})cos({5x-3x\over 2})\over 2cos ({5x+3x\over 2})cos({5x-3x\over 2})}\)
\(\left[\begin{array}{l} \because \sin C+\sin D=2 \sin \left(\frac{\mathbf{C}+\mathbf{D}}{2}\right) \cos \left(\frac{\mathbf{C}-\mathbf{D}}{2}\right) \\ \cos C+\cos \mathbf{D}=2 \cos \left(\frac{\mathbf{C}+\mathbf{D}}{2}\right) \cos \left(\frac{\mathbf{C}-\mathbf{D}}{2}\right) \end{array}\right]\)
\(={2 \sin \ x \cos \ x\over 2 \cos \ 4x \cos \ x}=\tan 4x\)
=R.H.S

According to the given information,
we have the following

In \(\Delta \)ABC, by cosine rule,
we have
\({ AB }^{ 2 }={ AC }^{ 2 }+{ BC }^{ 2 }-2AC.BC\quad cos\frac { \pi }{ 4 } \)
\(\therefore \quad AB=\sqrt { { (250) }^{ 2 }+{ (300) }^{ 2 }-2\times 250\times 300\times \frac { 1 }{ \sqrt { 2 } } } \)
\(=\sqrt { 62500+90000-75000\sqrt { 2 } } \)
\(=\sqrt { 152500-75000\times 1.44 } \)
\(=\sqrt { 152500-108000 } =\sqrt { 44500 } =210.95m\)

\(\theta =31'=\left( \frac { 31 }{ 60 } \times \frac { \pi }{ 180 } \right) rad\quad and\quad r=360000km\)
\(\because \quad \theta =\frac { l }{ r } \therefore \frac { 31 }{ 60 } \times \frac { \pi }{ 180 } =\frac { 1 }{ 360000 } \)
 Ans. 3247.62 km

(sin2x + sin6x)- sin4x=0
\(\Rightarrow\) 2sin4xcos2x-sin4x=0
\(\Rightarrow\) sin4x=0 or cos2x=\(\frac { 1 }{ 2 } \)
Ans
\(x=\frac { n\pi }{ 4 } \quad or\quad n\pi \pm \frac { \pi }{ 6 } \)

\(cos\quad A\quad =\quad \sqrt { 1-\frac { 9 }{ 25 } } =\frac { 4 }{ 5 } ,\)
\(sin\quad B=\quad -\sqrt { 1-\frac { 144 }{ 169 } } =-\frac { 5 }{ 13 }\)
\(cos(A+B)\quad =\quad -\frac { 33 }{ 65 } \)

\(cos\quad A\quad =\quad \sqrt { 1-\frac { 9 }{ 25 } } =\frac { 4 }{ 5 } ,\)
\(sin\quad B=\quad -\sqrt { 1-\frac { 144 }{ 169 } } =-\frac { 5 }{ 13 } \)
\(sin(A-B)\quad =\quad -\frac { 16 }{ 65 } \)

cos 6\(\theta\) + 6 cos 4\(\theta\) + 15 cos 2\(\theta\) + 10
= (cos 6\(\theta\) + cos 4\(\theta\)) + (5 cos 4\(\theta\) + 5 cos 2\(\theta\)) + (10 cos 2\(\theta\) + 10)
= (cos 6\(\theta\) + cos 4\(\theta\)) + 5 (cos 4\(\theta\) + cos 2\(\theta\)) + 10 (cos 2\(\theta\)+ 1)
= 2cos 5\(\theta\) cos \(\theta\) + 5 x 2 cos 3\(\theta\) cos \(\theta\)+ 10 x 2 cos \(\theta\) cos \(\theta\)
= 2cos \(\theta\) [cos5\(\theta\) + 5 cos 3\(\theta\) + 10 cos \(\theta\)]
\(\therefore {cos \theta + 6 cos 4 \theta + 15 cos 2 \theta + 10 \over cos 5 \theta + 5 cos 3 \theta + 10 cos \theta}=2cos \theta\)

L.H.S a sin\(\frac { A }{ 2 } \) .sin \(\frac { B-C }{ 2 } \)
= a sin\(\left( \frac { \pi }{ 2 } -\frac { B+C }{ 2 } \right) \).sin \(\left( \frac { B-C }{ 2 } \right) \)
= k sin A. cos \(\left( \frac { B+C }{ 2 } \right) \).sin \(\left( \frac { B-C }{ 2 } \right) \)
= k sin A. [sin B-sin C]
=  [sin A sin B-sin A sin c] ..(i)
Similarly,
b sin \(\frac { B }{ 2 } sin\left( \frac { C-A }{ 2 } \right) \)
=  \(\frac { k }{ 2 } \)[sin B sin C-sin B sin A]... (ii)
and C sin\(\frac { C }{ 2 } \) sin \(\left( \frac { A-B }{ 2 } \right) \)
=\(\frac { k }{ 2 } \) [sin C sin A-sin C sin B] ... (iii)
Adding (i), (ii) and (iii), we get
a sin \(\frac { A }{ 2 } \)sin\(\left( \frac { B-C }{ 2 } \right) \) +b sin\(\frac { B }{ 2 } \)sin \(\left( \frac { C-A }{ 2 } \right) \)+c sin \(\frac { C }{ 2 } \)sin \(\left( \frac { A-B }{ 2 } \right) \)
⇒ \(\frac { k }{ 2 } \) [sin A sin B-sin A sin C] + \(\frac { k }{ 2 } \)[sin B sin C-sin B sin A]+\(\frac { k }{ 2 } \) [sin C sin A-sin C sin B]
⇒ \(\frac { k }{ 2 } \)[sin A sin B-sin A sin C+sin B sin A-sin C sin A-sin C sin B]
⇒ \(\frac { k }{ 2 } \) [0]
⇒  0 R.H.S
Hence proved

Here cot x = \(3\over4\) tan x =\({1\over cot x}={4\over 3}\)
Now sec²x = 1 + tan² x
\(\Rightarrow sec^2x=1+({4\over3})^2\)
\(\Rightarrow sec^2x=1+{16\over 9}\)
\(\Rightarrow sec^2 x={25\over 9}\Rightarrow sec \ x= \pm {5\over3}\)
But x lies in third quadrant.
\(\therefore sec \ x ={-5\over 3}\)
\(cos x={1\over sec x}={-3\over5}\)
Also sin² x + cos²x= 1
\(\Rightarrow sin^2x+({-3\over 5})^2=1\)
\(\Rightarrow sin^2x=1-{9\over 25}\)
\(\Rightarrow sin^2x={16\over25}\Rightarrow sin x=\pm{4\over5}\)
But x lies in third quadrant.
\(\therefore sin \ x={-4\over 5}\)
\(\therefore cosec \ x={1\over sin \ x}={-5\over 4}\)