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Evaluate: 23(cos430°-sin445°)-3(sin260°-sec245°)+14cot230°. - Mathematics

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प्रश्न

Evaluate:

`2/3 (cos^4 30° - sin^4 45°) - 3(sin^2 60° - sec^2 45°) + 1/4 cot^2 30°`.

योग

उत्तर

`cos 30° = sqrt3/2, sin 60° = sqrt3/2, cot 30° = sqrt3, sin 45° = 1/sqrt2, sec 45° = sqrt2`   ...(I)

 

∴ `2/3 (cos^4 30° - sin^4 45°) - 3(sin^2 60° - sec^2 45°) + 1/4 cot^2 30°` ...(From I)

 

`= 2/3 [(sqrt3/2)^4 - (1/sqrt2)^4]- 3[(sqrt3/2)^2 - (sqrt2)^2] + 1/4 (sqrt3)^2`

 

`= 2/3 [9/16 - 1/4] - 3[3/4 - 2] + [1/4 × 3]`

 

`= 2/3 [9/16 - 4/16] - 3[3/4 - 8/4] + 3/4`

 

`= 2/3 [(9 - 4)/16] - 3[(3 - 8)/4] + 3/4`

 

`= 2/3 [5/16] - 3[(- 5)/4] + 3/4`

 

`= 2/3 × 5/16 - 3 × (- 5)/4 + 3/4`

 

`= 5/24 + 15/4 + 3/4`

 

`= (5 + 90 + 18)/24`

 

`= 113/24`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Trigonometric Ratios - Exercise 10.3 [पृष्ठ ५४]

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आरडी शर्मा Mathematics [English] Class 10
अध्याय 10 Trigonometric Ratios
Exercise 10.3 | Q 9.01 | पृष्ठ ५४

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