3cos²30 + sec²30 +2cos0 + 3sin90 – tan²60.Evaluate it.
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3cos²30° + sec²30° + 2cos0° + 3sin90° - tan²60°
we know, cos30° = √3/2
sec30° = 1/cos30° = 2/√3
cos0° = 1 , sin90° = 1
tan60° = √3
now, 3 × (√3/2)² + (2/√3)² + 2(1)² + 3(1) - (√3)²
= 3 × 3/4 + 4/3 + 2 + 3 - 3
= 9/4 + 4/3 + 2
= 9/4 + (4 + 6)/3
= 9/4 + 10/3
= (27 + 40)/12
= 67/12
we know, cos30° = √3/2
sec30° = 1/cos30° = 2/√3
cos0° = 1 , sin90° = 1
tan60° = √3
now, 3 × (√3/2)² + (2/√3)² + 2(1)² + 3(1) - (√3)²
= 3 × 3/4 + 4/3 + 2 + 3 - 3
= 9/4 + 4/3 + 2
= 9/4 + (4 + 6)/3
= 9/4 + 10/3
= (27 + 40)/12
= 67/12
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