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There can be no significant simplicfication if some of the angles are theta and others are x, so assume that all angles are x. [csc(x) - cot(x)]*[cos(x) + 1] =[1/sin(x) - cos(x)/sin(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos2(x)] =1/sin(x)*[sin2(x)] = sin(x)
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
whats the big doubt,cot/tan+1= 1+1= 2
For a start, try converting everything to sines and cosines.
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
1 cot(theta)=cos(theta)/sin(theta) cos(45 degrees)=sqrt(2)/2 AND sin(45 degrees)=sqrt(2)/2 cot(45 deg)=cos(45 deg)/sin(deg)=(sqrt(2)/2)/(sqrt(2)/2)=1
Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).
By converting everything to sines and cosines. Since tan x = sin x / cos x, in the cotangent, which is the reciprocal of the tangent: cot x = cos x / sin x. You can replace any other variable (like thetha) for the angle.
It is -sqrt(1 + cot^2 theta)
There can be no significant simplicfication if some of the angles are theta and others are x, so assume that all angles are x. [csc(x) - cot(x)]*[cos(x) + 1] =[1/sin(x) - cos(x)/sin(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos2(x)] =1/sin(x)*[sin2(x)] = sin(x)
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Yes, it is.
-2(cot2theta)
csc θ = 1/sin θ → sin θ = -1/4 cos² θ + sin² θ = 1 → cos θ = ± √(1 - sin² θ) = ± √(1 - ¼²) = ± √(1- 1/16) = ± √(15/16) = ± (√15)/4 In Quadrant III both cos and sin are negative → cos θ= -(√15)/4
It's 1/2 of sin(2 theta) .