In a Circle prove that `ar(ABC) = 1/2 r^2 sin theta`

In Delta ABC Prove that sin ^(2) "" (A)/(2) + sin ^(2) ""(B) /(2) + sin ^(2) ""( C) /(2) =1 -( r) / ( 2R)

In triangleABC , point D lies on side BC. E is the midpoint of AD. Prove that, ar(EBC)= 1/2 ar(ABC)

Prove that the area of a sector of circle with radius r and central angle theta^@ is 1/2 r^2 theta .

If ABCD is a parallelogram,the prove that ar(ABD)=ar(BCD)=ar(ABC)=ar(ACD)=1/2ar(||^(gm)ABCD)

Prove that (1) the area of a sector of circle with radius r and central angle theta^(@) is (1)/(2)r^(2)theta (ii) the area of a sector of a circle of radius r and an arc of length s is (1)/(2)rs

Prove that sin^(4) theta + cos^(4) theta = 1 - 2 sin^2 theta cos^(2) theta

Given that sin theta+2cos theta=1 then prove that 2sin theta-cos theta=2

If x=(2sin theta)/(1+cos theta+sin theta), then prove that (1-cos theta+sin theta)/(1+sin theta) is equal to x

Prove: (v) cot^2 theta-(1)/ (sin^2 theta) = -1