Question

If abcd is a cyclic quadrilateral hen the value of cos2a-cos2b-cos2c+cos2d

Answer 1

We know in a cyclic quadrilateral, a+c = 180° and also b+d = 180°

We have to find the value of cos2a-cos2b-cos2c+cos2d

Again by applying formulae of trigonometric identity

cos2a - cos2c = 2 sin (2a+2c/2) sin (2c-2a/2) = 2 sin (a+c) sin (c-a)........(1)

cos2d -cos2b = 2 sin (2b+2d/2) sin (2b-2d/2) = 2 sin (b+d) sin (b-d)........(2)

Now, we know that as the quadrilateral is cyclic so  (a+c) = 180°

and sin 180° is zero therefore eq. (1) becomes 0

Also sin(b+d) which is also equal to sin180° is zero, therefore eq. (2) also becomes 0.

Adding eq. (1) and (2) cos2a - cos2b - cos2c + cos 2d = 0

Hope this helps!!!

Answer 2

The ABCD quadrilateral are cyclic in nature that means that the sum of opposite angle is 180 degree

Hence, A + C = 180, B + D = 180

Cos 2a – Cos 2C = 2 Sin (2A + 2C/2) sin (2C – 2A/2) = 2 Sin (A + C) sin (C – A)

Cos 2D – Cs 2B = 2 Sin (2B +2D/2) Sin (2B-2D/2) = 2 Sin(B + D) Sin (B – D)

When you will inset A + C = 180 and B + D = 180 in the sin equation they will turn zero and then add both the equation,

cos2a - cos2b - cos2c + cos2d = 0