We know, sec(-x) = sec x
So, sec(-45°)=sec 45°=1/(cos 45°)=?2.

In 60 minutes, angle covered by minute hand is 360°.
In 30 minutes, angle covered by minute hand is 180°.
L=24 cm, ?=180°=?
?=L/r => ?=24/r => r=24/? = 38.2 cm.

According to law of sines, in any triangle ABC,
a/sinA = b/sinB = c/sinC
So, a/b = sinA/sinB
(a + b)/(a – b) = (sinA + sinB)/( sinA ? sinB)
=> (a + b)/(a – b) = (2 sin((a + b)/2) cos((a – b)/2))/ (2 sin((a + b)/2) sin((a – b)/2))
=> (a + b)/(a – b) = (tan(A + B)/2)/(tan(A – B)/2)
=> (tan(A + B)/2) = (a + b)/(a – b) (tan(A – B)/2)
=> (tan(π/2 + C/2)) = (a + b)/(a – b) (tan(A – B)/2)
=> cot (C/2) = (a + b)/(a – b) (tan(A – B)/2)
=> tan (C/2) = (a – b)/(a + b) (tan(A – B)/2)
Similarly, tan (A/2) = (b – c)/(b + c) (cot(B – C)/2).

sin (75°) = sin (45°+30°) = sin45° cos30° + cos45° sin30°
= (1/\(\sqrt{2}\) * \(\sqrt{3}\)/2) + (1/\(\sqrt{2}\) * 1/2) {sin(x + y)=sin x cos y + cos x sin y}
= (\(\sqrt{3}\) + 1)/2\(\sqrt{2}\).

Since sec x is negative in 2^nd and 3^rd quadrant so, re are two principal solutions to above equation i.e. x = π – π/3, π + π/3. So, x=2π/3, 4π/3.

sin 3x = sin (2x + x)
= sin 2x cos x + cos 2x sin x
= (2 sin x cos x) cos x + (1 ? 2sin² x) sin x
= 2 sin x (1 ? sin² x) + sin x ? 2 sin³ x
= 2 sin x ? 2 sin³ x + sin x ? 2 sin³ x
= 3 sin x ? 4 sin³ x.

We know, 2 sin x sin y = cos (x ? y) – cos (x + y)
2 sin 75° sin15° = cos (75°-15°) ? cos (75°+15°)
= cos 60° ? cos 90° = 1/2 – 0 = 1/2.

We know, tan x is not defined when cos x = 0.
cos x = 0 whenever x is π/2, 3π/2, 5π/2, ???? i.e. all odd integral multiples of π/2
so, x=(2n+1) π/2.

180 degree = ? radian
1 degree = ?/180 radians
60 degrees = 60* ?/180 radians = ?/3 radians
Angle=Arc length/Radius
?/3 = 20/Radius => Radius = 60/? = 19.09 cm.

Since tan x is negative in 2^nd and 4^th quadrant so, re are two principal solutions to above equation i.e. x = π – π/3, 2π – π/3. So, x=2π/3, 5π/3.