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If alpha=tan^(-1)((4x-4x^3)/(1-6x^2+x^4)...

If `alpha=tan^(-1)((4x-4x^3)/(1-6x^2+x^4)),beta=2sin^(-1)((2x)/(1+x^2))` and `tanpi/8=k ,` then (a) `alpha+beta=pi` for `x in [1,1/k]` (b)`alpha+beta` for `x in (-k , k)` (c) `alpha+beta=pi` for`x in [1,1/k]` (d) `alpha+beta=0` for `x in [-k , k]`

A

`alpha + beta = pi " for " x in [1, (1)/(k))`

B

`alpha = beta " for " x in (-k, k)`

C

`alpha + beta = - pi " for " x in [1, (1)/(k))`

D

`alpha + beta = 0 " for " x in (-k, k)`

Text Solution

Verified by Experts

The correct Answer is:
A, B
Put `x = tan theta`
`rArr alpha = tan^(-1) (tan 4 theta)`
`= 4 theta - pi " for " x in [1, (1)/(k))`
`= 4 theta " for " x in (-k, k)`
Also, `beta - 2(pi - 2 theta) " for " x in [1, (1)/(k))`
`= 4 theta " for " x in (-k, k)`
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