if tanA=2. evaluate secAsinA + tan2A-cosecA

it is 0

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its answer is 5....

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 it is 6 -(root 5)/2

secA sinA =sinA / cos A =tan A =2

tan2A=22 =4

cosec A is 1 / sinA and when we draw the triangle it comes root5 / 2

so according to the question my answer is right...

hope it helps!!

cheers!!!

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hope it helps!!

cheers!!

ay!

it is 

tan2A = 22 =4

you plz correct that...

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 thumbs up plzzz...

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 your answer is wrong

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 secA.sina =  2    

tan2A = 4    

1+ sec2A = 4

1/cos2A = 3 

sin2A =  4/3 = 2/sqrt3

Ans is            6sqrt3 - 2/ sqrt3

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 1+tan2A=sec2A

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no i am right.

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SecAsinA +  tan2A - CosecA

tanA + 4 -CosecA

2 + 4 - cosecA

6 - cosecA ------ 1

1 + Sec2A = tan2A

Sec2A = 4-1

1/cos2A = 3 

cos2A = 1/3

Sin2A/cos2A = 4

Sin2A X 1/cos2A = 4

3Sin2A = 4

Sin2A = 4/3

Sin A =  2/sqrt3 ---------- 2

From 1 & 2

you'll get the answer

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hey!!  Pradeep17Ch is right!!

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thnx pradeep

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 Can you make out any error in my calculation?

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 i am right....!

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  • 12- ROOT 5/2
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TanA=2 find secAsinA +tan2A-cosecA
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