If sinA =1//2 find the value of 2secA/1+TAN^2A
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sin A = 1/2
cos A = √3 / 2
Tan A = 1/3 √3
sec A = 1/cos A
sec A = 1 / √3/2
sec A = 2/3 √3
2sec A / 1 + tan ² A
= 2 (2/3) √3 / 1 + (1/√3)²
= 4√3 /3 / 1 + 1 / 3
= 4√3 / 2
cos A = √3 / 2
Tan A = 1/3 √3
sec A = 1/cos A
sec A = 1 / √3/2
sec A = 2/3 √3
2sec A / 1 + tan ² A
= 2 (2/3) √3 / 1 + (1/√3)²
= 4√3 /3 / 1 + 1 / 3
= 4√3 / 2
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HOW DID 3/ROOT3 BECOME 3 ROOT 3/3
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