URGENT!!

cos-1(x2-1 / x2+1) + 1/2 * tan-1(2x/1-x2) = 2Ï€/3.
Solve for x

28 Answers

1
Philip Calvert ·

yes prajith.....
"hey Philip

cos-1(cos(-2θ)=cos-1(cos(π-2θ).........dont u remember........"

toh maine kya log 10 likha hai iski jagah[11]
that pi - 2θ did not drop from the heavens

1
sriraghav ·

Thank u guys...

1
sriraghav ·

ok, i got it

11
Mani Pal Singh ·

pata nahin
i
didnt
saw ur
solution

and its different 4m ur ways(i hope )

1
MATRIX ·

[9][9][9]........thats wat the answer we got naa.........

11
Mani Pal Singh ·

here it is
i made a small mistake
and now corrected it!!!!!!!!!!!!!

11
Mani Pal Singh ·

k

1
sriraghav ·

Manipal where is ur post!!!!

1
sriraghav ·

@manipal. but wat abt using cos-1.....Where is the mistake in our method...

1
sriraghav ·

Its my mistake...Philip u r rite....( but my max sir hav specifically rounded the answer with red ink and wrote as wrong one)

1
sriraghav ·

hmm..... oK Thanks guys 4 ur efforts!!...I shud hav asked my teacher regarding this one one that day itself....The probability of this type of question in Inverse trigs is very very less(hoping)..Bye :-) ALL THE BEST 4 UR MAX xams..

1
Philip Calvert ·

see raghav .......maybe u got it wrong bcos of you wrong treatment of
cos-1(-x)

1
MATRIX ·

no philip i was just reminding u of that thing....thts all.i was hw our procedure can be wrong.........

1
MATRIX ·

yes Sriraghav.....im sure.........

1
Philip Calvert ·

put x= tan θ

u'll get

cos-1(-cos2θ) + θ = 2π/3
π - 2θ + θ = 2π/3
θ = π/3

x = √3

i hope it is correct...
though i have to admit i may have made some mistakes cuz didnt use pen and paper

1
sriraghav ·

Question is absolutely correct!!Guys r u sure root 3 is correct??

1
MATRIX ·

hey Philip

cos-1(cos(-2θ)=cos-1(cos(π-2θ).........dont u remember........

1
Philip Calvert ·

are u sure answer is not root 3 and the question is perfectly correct......
bcos most times this way we shud get the answer.....
i think that tan term in my solution has a mistake but cos-1 is perfectly correct........
___________________

1
MATRIX ·

sorry.......i didnt read it carefully.......

1
Philip Calvert ·

yes i have a feeling that root 3 might be wrong.. i am still trying

1
Philip Calvert ·

matrix pls read question carefully

its x sq -1 not 1- x sq in cos inverse term

1
sriraghav ·

post the full solution please....

1
MATRIX ·

manipal mistyped edited nw,.............

11
Mani Pal Singh ·

yeh integration kyun ho rahi hai?????

1
MATRIX ·

then after applying x=tanθ

cos-1(-(1-tan2θ)/(1+tan2θ))+tan-12tanθ/(1+tan2θ)

=cos-1(-cos2θ)+tan-1(tan2θ)

........ Rest is urs........now.........

11
Mani Pal Singh ·

bhai a better idea

cos-1x ko tan-1x ki form mein likh le and apply the formula
u will get the answer
if any troubles then ask again
i will tell

now i am taking dinner

1
Philip Calvert ·

wat root 3 is wrong .....[7]
then i have missed something ... ok lets see [12]

1
sriraghav ·

K philip.I did the same in my exam. my problem is cos-1(x2-1/x2+1)... I wrote cos(-θ) = cos(θ).. and got it as √3...But the answer is wrong... It is our pre board question.....Someone rectify if it is wrong...

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