Question

1.If sin2A = k then Cosec2A + cos2A.cot2A
2.If sin2A = k Then Sin6A
3.If sin A + sin^2 A + sin^3 A = 1 then cos^6 A– 4cos^4 A + 8cos^2 A is equal to :
4.The value of √3 cosec 20°- sec 20°

Solve these questions needed urgently​

Answer 1

Given : √3 cosec 20°- sec 20°

To Find : Value

Solution:

√3 cosec 20°- sec 20°

use Cosecx = 1/sinx   & secx = 1/Cosx

=  √3 /sin20°- 1/cos20°

=  (√3cos20°  - sin20°)/sin20°cos20°

Multiply and divide by 2

=  2 (√3cos20°  - sin20°)/2sin20°cos20°

Use Sin2x = 2sinx cosx

= 2  (√3cos20°  - sin20°)/sin40°

= 2  (2(√3/2)cos20°  - 2.(1/2)sin20°)/sin40°

= 2 * 2  ( (√3/2)cos20°  -  (1/2)sin20°)/sin40°

√3/2  =Sin60°

1/2 = Cos60°

= 4   ( Sin60°cos20°  -  Cos60°sin20°)/sin40°

Sin(A - B) = SinA CosB - CosASinB

= 4 Sin(60° - 20°) /sin40°

= 4 sin40°/sin40°

= 4

√3 cosec 20°- sec 20°   = 4

Please post Question one by one

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