Since B\(\left( \frac { 5 }{ 7 } ,\frac { 2\sqrt { 6 } }{ 7 } \right) \) is the point on the terminal side of an angle θ

OB² = OA² + AB²
= \((\frac{5}{7})^2+(\frac{2\sqrt{6}}{7})^2=\frac{25}{49}+\frac{24}{49}=\frac{49}{49}=1\)
ஃ ⇒ OB = 1
sin θ = \(\frac{opp}{hyp}=\frac{AB}{OB}=\frac{2\sqrt{6}}{7}\) , cos θ = \(\frac{adj}{hyp}=\frac{OA}{OB}=\frac{5}{7}\)
tan θ = \(\frac{opp}{adj}=\frac{\frac{2\sqrt{6}}{7}}{\frac{5}{7}}=\frac{2\sqrt{6}}{7}\times\frac{7}{5}=\frac{2\sqrt{6}}{5}\)
tan θ = \(\frac{\frac{2\sqrt{6}}{7}}{\frac{5}{7}}=\frac{2\sqrt{6}}{7}\times\frac{7}{5}=\frac{2\sqrt{6}}{5}\)
cot θ = \(\frac{5}{2\sqrt{6}}\)
cosec θ = \(\frac{1}{sinθ}=\frac{7}{2\sqrt{6}}\) ; sec θ = \(\frac{1}{cosθ}=\frac{7}{5}\)
and cot θ = \(\frac{1}{tanθ}=\frac{5}{2\sqrt{6}}\)

Given tan x = \(\frac{n}{n+1}\) and tan y = \(\frac{1}{2n+1}\)
\(\tan { \left( x+y \right) } =\frac { \tan { x } +\tan { y } }{ 1-\tan { x } ,\tan { y } } \)
\(=\frac { { 2n }^{ 2 }+2n+1 }{ { 2n }^{ 2 }+n+2n+1-n } =\frac { { 2n }^{ 2 }+2n+1 }{ { 2n }^{ 2 }+2n+1 } =1\)
 \(\therefore\tan { \left( x+y \right) } =1\)

Let r be the radius of the circle and  be the sector angle. the perimeter of the sector =  l + 2r = r\(\theta\) +2r
Length of the arc of a semi-circle of radius r is \(\pi \)r, Given that 2r + r\(\theta\) = \(\pi \)r = 2 +\(\theta\) = \(\pi \)
\(\Rightarrow \) θ = \((\pi -2)\) radians \((\pi -2)x\frac { 180 }{ \pi } =180-\frac { 360 }{ \pi } \)
= 180-114⁰32'44'
= 65⁰27^'16'' 

Radius of the circle = 100 cm
Length of arc  = 11 cm
∴ Let \(\theta \) radians be the angle subtended by the arc the center of the circle
Then \(\theta =\frac { l }{ r } =\frac { 22 }{ 100 } \)
We know \(\pi\) radians = 180⁰
1 radian = \(\left( \frac { 180 }{ \pi } \right) \)
\(\theta \) =\(180\times \frac { 7 }{ 22 } \times \frac { 22 }{ 100 } \quad \)
\(\theta \) = \(\frac { 18\times 7 }{ 10 } \)
\(\theta \) = \(\frac { 9\times 7 }{ 5 } =\frac { 63 }{ 5 } =12^o \frac { 3 }{ 5 } \)
\(\theta \) = \({ 12 }^{ 0 }36'\left[ \frac { { 3 }^{ 0 } }{ 5 } =\frac { 3 }{ 5 } \times 60=36^{ ' } \right] \)

Length covered in 1 hr by the train = 66 km

Length covered in (60 x 60) sec = 66 km
ஃ Length covered in 1 sec = \(\frac{66\times1000}{60\times60}\)m
Length covered in 20 sec l = \(\frac{1100}{3}\)m
ஃ Angle made in 20 sec = \(\frac{l}{r}=\frac{1100}{3\times1500}\)radians
ஃ θ = \((\frac{11}{45}\times\frac{180}{22}\times7)\) radians
θ = 14°

Given diameter of the circle is 40 cm
r = 20cm

Let  AB = 20 cm be a chord of the circle
Since OA = OB = AB = 20 cm, ΔAOB is equilateral
ஃ θ = ㄥAOB = 60°= 60 \(\times\) \(\frac{\pi}{180}=\frac{\pi}{3}\) radians
Let I be the length of the minor arc of the chord AB.
Then θ = \(\frac{l}{r}\) ⇒ l = rθ ⇒ l = 20(\(\frac{\pi}{3}\))
l = \(\frac{20\pi}{3}\) cm = 20 \(\times\) \(\frac{22}{7}\times\frac{1}{3}\) = 20.95 cm (app)

Given sin θ = \(\frac{1}{25}\)
\(cos\theta =\sqrt { 1-{ sin }^{ 2 }\theta } =\sqrt { 1-\frac { 1 }{ 625 } } =\frac { \sqrt { 624 } }{ 25 } =\frac { 4\sqrt { 39 } }{ 25 } \)
Now, \(sin\frac { \theta }{ 2 } =\sqrt { \frac { 1-cos\theta }{ 2 } } =\sqrt { \frac { 1-\frac { 4\sqrt { 39 } }{ 25 } }{ 2 } } =\sqrt { \frac { 25-4\sqrt { 39 } }{ 50 } } \)
\(cos\frac { \theta }{ 2 } =\sqrt { \frac { 1+cos\theta }{ 2 } } =\sqrt { \frac { 1-\frac { 4\sqrt { 39 } }{ 25 } }{ 2 } =\sqrt { \frac { 25-4\sqrt { 39 } }{ 50 } } } \)
Consider \(sin\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) =sin\frac { \pi }{ 4 } cos\frac { \theta }{ 2 } -cos\frac { \pi }{ 4 } sin\frac { \theta }{ 2 } \)
\(sin\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) =\frac { 1 }{ \sqrt { 2 } } cos\frac { \theta }{ 2 } -\frac { 1 }{ \sqrt { 2 } } sin\frac { \theta }{ 2 } \)
\(\frac { 1 }{ \sqrt { 2 } } \left( cos\frac { \theta }{ 2 } -sin\frac { \theta }{ 2 } \right) =\frac { 1 }{ \sqrt { 2 } } \left[ \sqrt { \frac { 25+4\sqrt { 39 } }{ 50 } } -\sqrt { \frac { 25-4\sqrt { 39 } }{ 50 } } \right] \)
\(=\frac { 1 }{ 5\sqrt { 2 } } \left[ \sqrt { \frac { 25+4\sqrt { 39 } }{ 50 } } -\sqrt { \frac { 25-4\sqrt { 39 } }{ 50 } } \right] \)

Given θ + Φ = \(\alpha\) and tan θ = k tan Φ
an θ = k tan Φ
\(\frac { tan\theta }{ tan\phi } =k\Rightarrow \frac { sin\theta cos\phi }{ cos\theta .sin\phi } =\frac { k }{ 1 } \)
⇒ \(\frac { sin\theta cos\phi }{ cos\theta .sin\phi } =\frac { k }{ 1 } \)
(By componendo and dividends)
⇒ ​​​​​\(\frac { sin\theta cos\phi -cos\theta sin\phi }{ cos\theta .sin\phi +cos\theta sin\phi } =\frac { k-1 }{ k+1 } \)
⇒  \(\frac { sin\left( \theta -\phi \right) }{ sin\left( \theta +\phi \right) } =\frac { k-1 }{ k+1 } \)
⇒  \(\frac { sin\left( \theta -\phi \right) }{ sin\alpha } =\frac { k-1 }{ k+1 } \)
⇒  \(sin\left( \theta -\phi \right) =\frac { k-1 }{ k+1 } sin\alpha \)

sin² (A + B) - sin² (A - B) = sin 2A sin 2B
LHS = sin² (A + B) - sin² (A - B)
= sin (A + B + A - B) sin (A +B - A + B)
= sin (2A), sin (2B) = RHS
Hence proved.

LHS = cos (A + B) cos C - cos (B + C) cos A
= (cos A cos B - sin A sin B) cos C - cos A (cos B cos C - sin B sin C)
= cos A cos B cos C - sin A sin B cos C - cos A cos B cos C + cos A sin B sin C
= cos A sin B sin C - sin A sin B cos C
= sin B (cos A sin C - sin A cos C)
= sin B (sin C cos A - cos C sin A)
= sin B sin (C - A)
= RHS
Hence proved.