Inverse Trigo (Help!)

Q . Solve:

cos-1 x2-1x2+1 + tan-1 2xx2-1 = 2Ï€3

I'm getting ±√3 but the book says √3, 2 - √3! [7]

12 Answers

11
Khilen Khara ·

I am getting √3 and -1/√3 as the answer..:)

30
Ashish Kothari ·

can you show your working here?

11
Khilen Khara ·

2tan-1 2xx2-1=2pi3
2xx2-1=√3
Therefore x=√3 and -13..:)

11
Khilen Khara ·

@ashish-u show me your working..

30
Ashish Kothari ·

My working was similar to yours except that I took,

2cos-1 x2-1x2+1 = 2Ï€3

x2-1x2+1 = 12

Giving, x= ±√3

262
Aditya Bhutra ·

you are making mistake guys.!!!

cos-1 1-x21+x2 = tan-12x1-x2

but in the question it is cos-1x2-11+x2.

30
Ashish Kothari ·

Now, isn't θ = cos-1 x2-1x2+1 = tan-1 2xx2-1 ?

262
Aditya Bhutra ·

just replace x=tanθ2. then cos-1x2-1x2+1 = cos-1 sin2θ/2 - cos2θ/21 = cos-1 (-cosθ) = π -θ

while tan-1 2x1-x2 = θ

1
chintan patel patel ·

here x2 should be<1
so x2-1<0
tan-12xx2-1<0
then i am confused what to do after this
aisa solve hoga kya?

30
Ashish Kothari ·

@Aditya - All said and done, but how are we getting the given answer? I might be missing something so please if you can explain. :)

@Nishant sir- Please tell us where we are going wrong!

262
Aditya Bhutra ·

the answer in the book must be surely wrong. bcoz if u proceed with my method as i have given before, we get π -2θ = π/3 (where θ=2tan-1x )

therefore x= 1/√3 , √3

1
Shankar C ·

the book ans may be correct........
i got 2-√3

let x= tanθ

cos-1 (-cos 2θ) + tan-1(-tan 2θ) = 2pi/3
pi -2θ-2θ = 2pi/3

θ=pi/12
tan pi/12 = 2 - √3

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