Given A + B + C = \(\frac{\pi}{2}\) \(\Rightarrow\) cos (A + B + C) = cos \(\frac{\pi}{2}\)
\(\Rightarrow\) cos (A + B + C) = 0
cos [A + B + C] = cos (A + B) cos C - sin (A + B) sin C
= cos C [cos A cos B - sin A sin B] - sin C [sin A . cos B + cos A sin B]
= cos A cos B cos C - sin A sin B cos C - sin A cos B sin C - cos A sin B sin C
0 = cos A cos B cos C - sin A sin B cos C - sin A cos B sin C - cos A sin B sin C
cos A cos B cos C = sin A sin B cos C + sin A cos B sin C + cos A sin B sin C
Hence proved.

LHS = \(sin^{ 2 }\frac { 3\pi }{ 4 } +2cos^{ 2 }\frac { \pi }{ 4 } +2sec^{ 2 }\frac { \pi }{ 3 } \)=\(2\left( sin\frac { 3\pi }{ 4 } \right) ^{ 2 }+2\left( cos\frac { \pi }{ 4 } \right) ^{ 2 }+2\left( sec\frac { \pi }{ 3 } \right) ^{ 2 }\)
= \(2\left( sin\frac { \pi }{ 4 } \right) ^{ 2 }+2\left( cos\frac { \pi }{ 4 } \right) ^{ 2 }+2\left( sec\frac { \pi }{ 3 } \right) ^{ 2 }\)   \(\left[ \because sin\frac { 3\pi }{ 4 } =sin(\pi -\frac { \pi }{ 4 } )=sin\frac { \pi }{ 4 } \right] \)
= \(2\left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }+2\left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }+{ 2(2) }^{ 2 }\) = \(2\left( \frac { 1 }{ 2 } \right) +2\left( \frac { 1 }{ 2 } \right) +2(4)\)
= 1 + 1 + 8 =10 = RHS

Given sec x =\(\sqrt { 2 } \)  ⇒ cos x = \(\frac { 1 }{ \sqrt { 2 } } \)
Since x lines in the IV quadrant, sinx is negative
\(sinx=-\sqrt { 1-cos^{ 2 }x } =-\sqrt { 1-\frac { 1 }{ 2 } } =-\sqrt { \frac { 1 }{ 2 } } =\frac { -1 }{ \sqrt { 2 } } \)
∴ \(cosecx=\frac { 1 }{ sinx } =-\sqrt { 2 } \) and tan\(x=\frac { sinx }{ cosx } =\frac { \frac { -1 }{ \sqrt { 2 } } }{ \frac { 1 }{ \sqrt { 2 } } } =-1\) and cot x = \(\frac { 1 }{ tanx } =-1\)
\(\frac { 1+tanx+cosecx }{ 1+cotx-cosecx } \) = \(\frac { 1-1-\sqrt { 2 } }{ 1-1+\sqrt { 2 } } =\frac { -\sqrt { 2 } }{ -\sqrt { 2 } } \) = -1

LHS = 1 + cos 2x + cos 4x + cos 6x
= ( cos 0x + cos 2x ) + ( cos 4x + cos 6x )
\(=\left[ 2 cos \left( {{0+2}\over{2}} \right)x.cos \left({{2x-x}\over{2}} \right)\right]+\left[ 2\ cos\left({{4x+6x}\over{2}} \right).cos\left( {{6x-4x}\over{2}} \right) \right]\left[ \because cos\ C+cos\ D=2cos\left( {{C+D}\over{2}} \right) .cos\left( {{C-D}\over{2}} \right)\right]\)
= 2 cos x. cos x + 2 cos 5x. cos x
= 2 cos x ( cos x + cos 5x )
= 2 cos x . \(\left( 2\ cos\left({{5x+x}\over{2}} \right) .cos\left( {{5x-x}\over{2}} \right)\right) \)
= 2 cos x.2.cos 3x. cos 2x
= 4 cos x cos 2x cos 3x = RHS.
Hence proved.

LHs = 4 cos 12° cos 48° cos 72°
= 2 ( 2 cos 12° cos 48°) cos 72°
= 2 ( cos 60° + cos 36° ) cos 72°
= 2 cos 60° cis 72° + 2 cos 36° cos 72°    [ \(\because\) 2 cos A cos B = cos ( A + B ) + cos ( A - B ) ]
= \(2\times{{1}\over{2}}\) cos 72° + 2 cos 36° cos 72°
= cos 72° + 2 cos 36° cos 72°
= cos 72° + cos (108°) + cos 36°   [ \(\because\) 2 cos A cos B = cos ( A + B ) + cos ( A - B ) ]
= cos 72° + cos (108-72°) + cos 36°
= cos 72°-cos 72°+cos 36°  [ \(\because\) cos (180-\(\theta\)) =-cos \(\theta\)]
= cos 36°
= RHS
Hence, proved.

LHS = sin⁸x - cos⁸x
= (sin⁴x)²- (cos⁴x)²
= (sin⁴x - cos⁴x) (sin⁴x + cos⁴x)
= (sin²x + cos²x) (sin²x - cos²x) (sin⁴x + cos⁴x) = (sin²x - cos²x) (sin⁴x + cos⁴x)   [∵ sin²x + cos²x = 1]
= (sin²x - cos²x) (sin⁴x + cos⁴x + 2 sin²x cos²x - 2sin²x cos²x)   (Adding and subtracting 2sin²x cos²x)
= (sin²x - cos²x) ((sin²x + cos²x)²- 2sin²x cos²x)      [∵ (sin²x + cos²x)² = sin⁴x + cos⁴x + 2sin²x cos²x]     [∵ sin²x + cos²x = 1]
= (sin²x - cos²x) (1 - 2sin²x - cos²x)
= RHS

LHS = sin 75^o - sin 15^o
= sin (90 - 15^o) - sin 15^o
= cos 15^o - sin 15^o
RHS = cos 105^o + cos 15^o
= cos (90 + 15^o) + cos 15^o
= - sin 15^o + cos 15^o = cos 15^o - sin 15^o
\(\therefore\) LHS = RHS
Hence proved.

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

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)

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