(a} In the diagram, O is the centre of the circle ABCDE, = I\(\overline{BC}\)I = |\(\overline{CD}\)| and < BCD = 108°. Find < CDE. (b) Given that tan x = \(...

Answer Details

(a) To find < CDE, we need to first find < BOC.

Since O is the center of the circle, < BOC is twice the angle < BAC. Therefore,

< BAC = 1/2 * < BOC

= 1/2 * (360° - < BCD - < ABC)

= 1/2 * (360° - 108° - 36°)

= 1/2 * 216°

= 108°

Similarly, we can find that < BDC = 1/2 * (360° - < BCD - < CDE)

= 1/2 * (360° - 108° - < CDE)

= 126° - 1/2 * < CDE

Since < BDC and < BCD are equal, we have:

126° - 1/2 * < CDE = 108°

1/2 * < CDE = 18°

< CDE = 36°

Therefore, < CDE is 36°.

(b) We are given that tan x = √3, and we need to find the value of:

(cos x)^2 - sin x

---------------------

(sin x)^2 + cos x

Using the identity (cos x)^2 + (sin x)^2 = 1, we can write:

(cos x)^2 = 1 - (sin x)^2

Substituting this into the expression, we get:

[(1 - (sin x)^2) - sin x] / [(sin x)^2 + cos x]

Simplifying, we get:

(1 - 2(sin x)^2 - sin x) / [(sin x)^2 + cos x]

Substituting tan x = √3, we have:

sin x / cos x = √3

sin x = √3 cos x

Substituting this into the expression, we get:

(1 - 2(3cos^2 x) - √3 cos x) / (3cos^2 x + cos x)

Multiplying both the numerator and denominator by -1, we get:

(2(3cos^2 x) + √3 cos x - 1) / (3cos^2 x + cos x)

Simplifying, we get:

(6cos^2 x + √3 cos x - 1) / (3cos^2 x + cos x)

Substituting √3 for sin x / cos x, we have:

(6/4 + √3/4 - 1) / (3/4 + 1/4)

= (9/4 + √3/4) / 1

= 9 + √3

Therefore, the value of the expression is