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Trignometry

Blazing goIITian

Joined:	30 Aug 2011
Post:	359

15 Mar 2012 08:31:30 IST

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431

inverse trig

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Trigonometry

find the value of

32cosec^2( 0.5arctan 4/3)+27/2 sec^2(0.5 arctan3/4)

Comments (7)

sathyaram

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15 Mar 2012 09:36:19 IST

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32cosec^2( 0.5arctan 4/3)=160

sudhindra shenoy

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15 Mar 2012 13:01:01 IST

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ans is correct plzz give d explanation

hemang

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18 Mar 2012 00:36:45 IST

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first we calculate 0.5*tan inverse 4/3.

let tanx = 4/3.

so tan(2*x/2) = [2tan(x/2)]/[1 - tan²(x/2)] = 4/3.

we get tan(x/2) = 1/2 , -2.

cosec(x/2) = root5.

so we have 32*cosec^2(0.5 * arctan4/3) = 160.

similarly let tany = 3/4.

we want tan(y/2). we can calculate to get, tan(y/2) = -3, 1/3.

(27/2)*sec^2(0.5*arctan3/4) = 15..

sudhindra shenoy

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18 Mar 2012 08:03:06 IST

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oops the reqd answer is187. sorry 4 d wrong answer of 160

sudhindra shenoy

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18 Mar 2012 08:43:12 IST

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i did it by calculator i got 175 as the ans i dont know what d hell is d actual ans .In d book it is given as 187

atul

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19 Mar 2012 00:17:24 IST

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175 is correct answer

hemang

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20 Mar 2012 12:45:40 IST

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i am getting 175 again.

have done it twice now. :)

the calculator is correct, don't worry.

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