We know, sin 3x = 3sin x – 4sin³x
= 3(1/2) ? 4(1/2)³
= 3/2 ? 4/8 = 3/2 – 1/2 = 1.

We know, cosec²x ? cot²x = 1
So, cot²x ? cosec²x = -1.

Radius of circle=2 cm
In 60 seconds, angle covered by second hand is 360°
In 40 seconds, angle covered by second hand is 360°*4/6 = 240°
240°=240*
Angle=Arc length/Radius
240*?/180 = Arc length/2
Arc length=8 ?/3 = 12.56 cm.

We know, sin 2x = (2 tan x) / (1 + tan²x)
sin 2x=(2*1/\(\sqrt{3}\)) / (1+1/3) = (2*3)/4*\(\sqrt{3}\) = \(\sqrt{3}\)/2.

The expression involving integer ?n? which gives all solutions of a trigonometric equation is called general solution. n is used in general solution only.

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) (cot(A – B)/2).

sin x=-4/5
We know, sin²x + cos²x=1
cos²x = 1-(-4/5)² = 1-16/25=9/25
cos x=±3/5
cos x is negative in 3^rd quadrant so, cos x=-3/5.
sec x = 1/cos x = 1/(-3/5) = – 5/3.

We know, cos (-x) = cos x
So, cos(-60°) = cos 60°=1/2.

Since cosec x is positive in 1^st and 2^nd quadrant, so re are two principal solutions to above equation i.e. π/6, π – π/6. So, x=π/6, 5π/6.

tan 1560° = tan (360°*4 + 120°) = tan 120° {tan (2nπ+x) = tan x}
= tan (180°-60°) = -tan 60° = – ?3. {tan π-x = – tan x}