\(2cos\theta=x+\frac{1}{x}\)
\(\therefore 4cos^2\theta=(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2\)
\(4cos^2\theta-2=x^2+\frac{1}{x^2}\)
\(=2(2cos^2\theta-1)=x^2+\frac{1}{x^2}\)
\(\Rightarrow cos2\theta=\frac{1}{2}(x^2+\frac{1}{x^2})\)

LHS \(\begin{aligned} =\cos \frac{3 \pi}{4} \cos x-\sin \frac{3 \pi}{4} \sin x-\cos \frac{3 \pi}{4} \cos x-\sin \frac{3 \pi}{4} \sin x \end{aligned}\)
\(\begin{aligned} =-2 \sin \left(\pi-\frac{\pi}{4}\right) \sin x=-2\left(\frac{1}{\sqrt{2}}\right) \sin x=-\sqrt{2} \sin x \end{aligned}\)
 

Since \(\pi,x<{{3\pi}\over{2}},\) x lies in the III quadrant only cot x and tan x are positive
Also \(\pi,x<{{3\pi}\over{2}},\) y is also lies in the III quadrant
Only cot y and tan y are positive
 
\(sinx=-\frac { 3 }{ 5 } \quad siny=-\frac { 24 }{ 25 } \)
\(cosx=-\frac { 4 }{ 5 } \quad cosy=-\frac { 7 }{ 25 } \)
ஃ cos(x - y) = cosx cosy + sinx siny
\(=\left( -\frac { 4 }{ 5 } \right) \left( -\frac { 7 }{ 25 } \right) +\left( -\frac { 3 }{ 5 } \right) \left( -\frac { 24 }{ 25 } \right) \)
\(=\frac { 28 }{ 125 } +\frac { 72 }{ 125 } =\frac { 100 }{ 125 } =\frac { 4 }{ 5 } \)

Since 0\(\frac { \pi }{ 2 } \) and 0\(\frac { \pi }{ 2 } \), A, B lies in the I quadrant
ஃ All the trigonometric ratios are positive

=\(\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } =4\)

=\(\sqrt { { 41 }^{ 2 }-{ 9 }^{ 2 } } \)
=40
\(sinA=\frac { 3 }{ 5 } \quad sinB=\frac { 40 }{ 41 } \)
\(cosA=\frac { 4 }{ 5 } \quad cosB=\frac { 9 }{ 41 } \)
sin (A + B) = sinA cosB + cosA sinB
=\(\frac { 3 }{ 5 } .\frac { 9 }{ 41 } +\frac { 4 }{ 5 } .\frac { 40 }{ 41 } \)
\(=\frac { 27 }{ 205 } +\frac { 160 }{ 205 } =\frac { 187 }{ 205 } \)

cos A = \(\frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 2bc } \)
= \(\frac { 25+49-9 }{ 2(5)(7) } =\frac { 65 }{ 70 } =\frac { 13 }{ 14 } \)
cosB = \(\frac { { c }^{ 2 }+{ a }^{ 2 }-{ b }^{ 2 } }{ 2ca } =\frac { 49+9-25 }{ 2(7)(3) } =\frac { 33 }{ 42 } =\frac { 11 }{ 14 } \) .

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

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
=\(\left( \frac { 4 }{ 5 } \right) \left( \frac { 12 }{ 13 } \right) -\left( \frac { -3 }{ 5 } \right) \left( \frac { -5 }{ 13 } \right) =\frac { 48 }{ 65 } -\frac { 15 }{ 65 } =\frac { 33 }{ 65 } \). 

\(\sin { \left( \pi +\theta \right) } =-\sin { \theta } \)
\(\sin { \left( \pi +\theta \right) } =\sin { \pi } \cos { \theta } +\cos { \pi } \sin { \theta } \)
\(=\left( 0 \right) \cos { \theta } +\left( -1 \right) \sin { \theta } \)
\(=0-\sin { \theta } \)
\(\sin { \left( \pi +\theta \right) } =-\sin { \theta } \)

\(\tan { \left( \frac { 7\pi }{ 12 } \right) } =​​\tan { \left( \frac { 7\times 180 }{ 12 } \right) } =\tan { \left( { 105 }^{ o } \right) } =\frac { \sin { { 105 }^{ o } } }{ \cos { { 105 }^{ o } } } \)
\(=\frac { \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } }{ \frac { 1-\sqrt { 3 } }{ 2\sqrt { 2 } } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \times \frac { 2\sqrt { 2 } }{ 1-\sqrt { 3 } } =\frac { \sqrt { 3 } +1 }{ 1-\sqrt { 3 } } \)
\(=\frac { { \left( \sqrt { 3 } +1 \right) }^{ 2 } }{ { 1 }^{ 2 }-{ \left( \sqrt { 3 } \right) }^{ 2 } } =\frac { 3+1+2\sqrt { 3 } }{ 1-3 } =\frac { \sqrt { 3 } +1 }{ 1-\sqrt { 3 } } \times \frac { 1+\sqrt { 3 } }{ 1+\sqrt { 3 } } \)
\(=-\left( \frac { 4+2\sqrt { 3 } }{ 2 } \right) =-2\left( \frac { 2+\sqrt { 3 } }{ 2 } \right) \)
\(\tan { \left( \frac { 7\pi }{ 2 } \right) } =-\left( 2+\sqrt { 3 } \right) \)

Let sin^-1 \(\left( \frac { 12 }{ 13 } \right) +{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =\pi \)
Then \(sin\quad x=\frac { 12 }{ 13 } ,cosy\frac { 4 }{ 5 } ,\quad and\quad tanz\frac { 63 }{ 16 } \)
\(cos\quad x\sqrt { 1-{ sin }^{ 2 }x } =\sqrt { 1-\frac { 144 }{ 169 } } =\sqrt { \frac { 25 }{ 169 } } =\frac { 5 }{ 13 } \)
\(and\quad tan\quad x=\frac { sin\quad x }{ cos\quad x } =\frac { 12 }{ 13 } /\frac { 5 }{ 13 } =\frac { 12 }{ 5 } \)
\(sin\quad y\sqrt { 1-{ cos }^{ 2 }y } =\sqrt { 1-\frac { 16 }{ 25 } } =\sqrt { \frac { 9 }{ 25 } } =\frac { 3 }{ 5 } \)
\(tan\quad y=\frac { sin\quad y }{ cos\quad y } =\frac { \frac { 3 }{ 5 } }{ \frac { 4 }{ 5 } } =\frac { 3 }{ 4 } \)
i.e have tan (x + y) = \(\frac { tan\quad x+tan\quad y }{ 1-tan\quad x.tan\quad y } \)
\(\frac { \frac { 12 }{ 5 } +\frac { 3 }{ 4 } }{ 1-\frac { 12 }{ 5 } \times \frac { 3 }{ 4 } } =\frac { \frac { 48+15 }{ 20 } }{ \frac { 20-36 }{ 20 } } =-\frac { 63 }{ 16 } \)
From (1) and (2), tan (x+y) = -tan z
\(\Rightarrow tan\quad (x+y)=tan\quad (-z)\)
\(\Rightarrow tan(x+y)=tan(\pi -z)\)
\(\Rightarrow x+y= -z\quad or\quad x+y=\pi -z\)
Since x, y, and z are positive, x + y \(\neq \) -z
\(\therefore\)  x + y + z = p
\(\Rightarrow { sin }^{ -1 }\left( \frac { 12 }{ 13 } \right) +{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =\pi \)  

Given a = 4, b = 6 and c = 8.
By cosine formula, \(\cos { B } =\frac { { a }^{ 2 }+{ c }^{ 2 }-{ b }^{ 2 } }{ 2ac } \)
\(\Rightarrow \cos { B } =\frac { 16+64-36 }{ 2(4)(8) } =\frac { 80-36 }{ 64 } =\frac { 44 }{ 64 } =\frac { 11 }{ 16 } ....(1)\)
Also \(\cos { C } =\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab } =\frac { 16+36-64 }{ 2(4)(6) } \)
\(=\frac { 52-64 }{ 48 } =\frac { -12 }{ 48 } =\frac { -1 }{ 4 }.....(2)\)
Now, LHS = \(4\cos { B } +3\cos { C } \)
\(=4\left( \frac { 11 }{ 16 } \right) +3\left( \frac { -1 }{ 4 } \right) \)      [ From (1) and (2) ]
= \(\frac { 11 }{ 4 } -\frac { 3 }{ 4 } =\frac { 11-3 }{ 4 } =\frac { 8 }{ 4 } =2=RHS\)   
Hence Proved.

\(LHS=sin\frac { \theta }{ 2 } sin\frac { 7\theta }{ 2 } +sin\frac { 3\theta }{ 2 } sin\frac { 11\theta }{ 2 } \)
\(=\frac { 1 }{ 2 } \left[ cos\left( \frac { \theta }{ 2 } -\frac { 7\theta }{ 2 } \right) -cos\left( \frac { \theta }{ 2 } -\frac { 7\theta }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ cos\left( \frac { 3\theta }{ 2 } -\frac { 11\theta }{ 2 } \right) -cos\left( \frac { 3\theta }{ 2 } -\frac { 11\theta }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ cos(-3\theta )-cos(4\theta )+cos(-4\theta )-cos7\theta \right] \)
\(=\frac { 1 }{ 2 } \left[ cos3\theta -cos4\theta +cos4\theta -cos7\theta \right] \quad \quad \left[ \because cos(-\theta )=cos\theta \right] \)
\(=\frac { 1 }{ 2 } \left[ cos3\theta -cos7\theta \right] =\frac { 1 }{ 2 } \left[ 2sin\left( \frac { 3\theta +7\theta }{ 2 } \right) .sin\left( \frac { 7\theta -3\theta }{ 2 } \right) \right] \)
\(=sin5\theta .sin2\theta =RHS.\)

Given sin θ + cos θ = m
LHS = cos⁶ θ + sin⁶θ
= (cos²θ)³+ (sin² θ)³
= (cos²θ + sin²θ)(cos⁴θ - cos²θ sin²θ + sin⁴θ)
= 1(cos⁴θ - cos²θ sin²θ + sin⁴θ)
= (cos²θ)²+ (sin²θ)² - cos²θsin²θ
= (cos²θ)²+ (sin²θ)²- cos²θ sin²θ
= 1 - 3 sin²\(\theta\) cos²\(\theta\) ....(1)
RHS = \(\frac { 4-3{ \left( { m }^{ 2 }-1 \right) }^{ 2 } }{ 4 } \)
= \(\frac { 4-3{ \left[ { \left( sin\theta +cos\theta \right) }^{ 2 }-1 \right] }^{ 2 } }{ 4 } =\frac { 4-3\left[ { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta +2sin\theta cos\theta -1 \right] }{ 4 } \)
= \(\frac { 4-12{ sin }^{ 2 }\theta { cos }^{ 2 }\theta }{ 4 } =\frac { 4 }{ 4 } -\frac { 12 }{ 4 } { sin }^{ 2 }\theta { cos }^{ 2 }\theta \)
= 1 - 3 sin²\(\theta\) cos²\(\theta\) ...(2)
From (1) and (2), LHS = RHS

Given a cos \(\theta\) - b sin \(\theta\) = C
Squaring both sides we get, (a cos \(\theta\) - b sin \(\theta\))² = c²
\(\Rightarrow \) a²cos²\(\theta\) + b²sin²\(\theta\) - 2ab sin \(\theta\) cos \(\theta\) = c²
\(\Rightarrow \) a²(1 - sin²\(\theta\))+ b² (1-cos²\(\theta\)) - 2ab sin \(\theta\) cos \(\theta\) = c²
\(\Rightarrow \) a² - a² sin² \(\theta\) + b² - b² cos2 \(\theta\) - 2ab sin \(\theta\) cos \(\theta\) = c²
\(\Rightarrow \) -a² sin²\(\theta\) -b²cos²\(\theta\) - 2ab sin \(\theta\) cos \(\theta\) = c² -a² - b²
\(\Rightarrow \) a²sin²\(\theta\) + b²cos2 \(\theta\) + 2ab sin cos = a²+ b²- c²
\(\Rightarrow \) (a sin \(\theta\)+ b cos \(\theta\))² = a² +  b² - c²
\(\Rightarrow \) a sin \(\theta\) + b cos \(\theta\) = \(\pm \sqrt { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } } \)
Hence proved

\(\frac{\sin (A-B)}{\cos A \cos B} =\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B} \)
\(=\tan A-\tan B \)
\(\text { L.H.S } =\tan A-\tan B+\tan B-\tan C+\tan C-\tan A=0\)

\(\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots \ldots \ldots+\cos 179^{\circ} \)
\(=\left(\cos 1^{\circ}+\cos 179^{\circ}\right)+\left(\cos 2^{\circ}+\cos 178^{\circ}\right)+ ...\)
\(=2 \cos 90^{\circ} \cos 89^{\circ}+2 \cos 90^{\circ} \cos 80^{\circ}+\ldots \ldots . \)
\(=0+0+0 \ldots \ldots=0 \)

\(x =\frac{1}{\cos 80^{\circ}}-\frac{\sqrt{3}}{\sin 80^{\circ}} \)
\(=\frac{\sin 80^{\circ}-\sqrt{3} \cos 80^{\circ}}{\sin 80^{\circ} \cos 80^{\circ}} \)
\(\frac{x}{2} =\frac{\frac{1}{2} \sin 80^{\circ}-\frac{\sqrt{3}}{2} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 80^{\circ} \cos 60^{\circ}-\cos 80^{\circ} \sin 60^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 20^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 160^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{2 \sin 80^{\circ} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}}=4 \)