Question

Find the value of sin²Dcos²D (0° < D < 90°)if  

(cosD – secD)² – (cosecD – sinD)² = 0.6

1. 12/13
2. 5/18
3. 4/15
4. 10/9

Answer 1

Correct Answer - Option 2 : 5/18

Given:

(cosD – secD)² – (cosecD – sinD)² = 0.6

Concepts used:

secD = 1/cosD

cosecD = 1/sinD

sin²D + cos²D = 1

sin2D = 2 × sinD × cosD

sin6D + cos6D = 1 – 3sin2Dcos2D

Calculation:

(cosD – secD)² – (cosecD – sinD)² = 0.6

⇒ {cosD – (1/cosD)}² + {(1/sinD) – sinD}² = 0.6

⇒ {(cos²D – 1)²/cos²D} + {(1 – sin²D)²/sin²D} = 0.6 

⇒ {(1 – cos2D)2/cos2D} + {(1 – sin2D)2/sin2D} = 0.6 

⇒ (sin²D)²/cos²D + (cos²D)²/sin²D = 0.6 

⇒ sin⁴D/cos²D + cos⁴D/sin²D = 0.6 

⇒ (sin⁶D + cos⁶D)/sin²Dcos²D = 0.6 

⇒ (1 – 3sin²Dcos²D)/sin²Dcos²D = 0.6 

⇒  1 – 3sin2Dcos2D = 0.6sin2Dcos2D

⇒ 1 = 3.6sin2Dcos2D

⇒ sin²Dcos²D = 1/3.6 

⇒ sin²Dcos²D = 10/36 = 5/18

∴ The value of sin2Dcos2D is 5/18