Let r₁, r₂ be the radii of two circles at the centres of which arcs of equal lengths subtend angles of 60° and 75° respectively.

ஃ θ = 60°= \(60\times\frac{\pi}{180}=\frac{\pi}{3}\) radians
θ = \(\frac{l}{r_1}⇒l=θ_1r_1\)
l = \(\frac{\pi}{3}r_1\)...(1)
θ₂ = 75° = \(75\times\frac{\pi}{180}=\frac{15\pi}{36}=\frac{5\pi}{12}\)radian
θ₂ = \(\frac{l}{r_2}⇒l=θ_2r_2\)
l = \(\frac{5\pi}{6}\times r_2\) ...(2)
From (1) and (2), \(\frac{\pi}{3}r_1=\frac{5\pi}{12}r_2\)
\(\frac{r_1}{r_2}=\frac{5\pi}{12}\times\frac{3}{\pi}=\frac{5}{4}\)
r₁ : r₂ = 5 : 4

Since the athlete runs 5 times around the circle, the distance covered = 5 (circumference of circle)
But given distance covered = 1 km

ஃ 5(2πr) =1 km = 1000m
⇒ \(10\times\frac{22}{7}\times r=1000\)
⇒ \(\frac{1000\times7}{10\times22}=\frac{100\times7}{22}=\frac{50\times7}{11}=\frac{350}{11}\)
r = 31.818m
r = 31.82 m

\(LHS=\frac { sin4x+sin2x }{ cos4x+cos2x } \)
\(=\frac { 2sin\left( \frac { 4x+2x }{ 2 } \right) .cos\left( \frac { 4x-2x }{ 2 } \right) }{ 2cos\left( \frac { 4x+2x }{ 2 } \right) .cos\left( \frac { 4x-2x }{ 2 } \right) } =\frac { sin3x.cosx }{ cos3x.cosx } =tan3x\) = RHS

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

395⁰
395⁰ = 360⁰ + 35⁰
\(\Rightarrow \) 395 - 35⁰ = 360⁰
∴ Coterminal angle For 395⁰ is 35⁰ 

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 < x<\(\frac{\pi}{2}\) and y = 0\(\frac{\pi}{2}\) , x and y lies in the I quadrant.
ஃ All the trigonometric ratios are positive.
\(\sqrt { { 17 }^{ 2 }-{ 15 }^{ 2 } }\) \(\sqrt { { 13 }^{ 2 }-12^{ 2 } } \)
\(\sqrt { 289-225 } \) = 5
= \(\sqrt { 64 } \)
= 8

\(sin\quad x=\frac { 15 }{ 17 } \quad sin\quad y=\frac { 5 }{ 13 } \)
\(cos\quad x=\frac { 8 }{ 17 } \quad cos\quad y=\frac { 12 }{ 13 } \)
\(tan\quad x=\frac { 15 }{ 8 } \quad tan\quad y=\frac { 5 }{ 12 } \)
= sin x cos y + cos x sin y
=\(\frac { 15 }{ 17 } +\frac { 12 }{ 13 } +\frac { 8 }{ 17 } .\frac { 5 }{ 13 } \)
=\(\frac { 180 }{ 221 } +\frac { 40 }{ 221 } =\frac { 220 }{ 221 } \)

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

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