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For the equation 1-2x-x^2=tan^2(x+y)+cot...

For the equation `1-2x-x^2=tan^2(x+y)+cot^2(x+y)` exactly one value of `x` exists exactly two values of `x` exists `y=-1+npi+pi/4,n in Z` `y=1+npi+pi/4, n in Z`

A

exactly one values of x exists

B

exactly two values of x exists

C

`y=-1+n pi+pi//4, n in Z`

D

`y=1+npi+pi//4, n in Z`

Text Solution

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The correct Answer is:
A, D

`1-2x-x^(2)=tan^(2) (x+y)+cot^(2) (x+y)`
`rArr -(x+1)^(2)=[tan (x+y)-cot(x+y)]^(2)`
Now L.H.S. `le 0` and R.H.S. `ge 0`
`rArr -(x+1)^(2)=[tan(x+y)-cot (x+y)]^(2)=0`
`rArr x=-1` and `tan (x+y)=cot (x+y)`
`rArr x=-1` and `tan^(2) (-1+y)=1`
`rArr x=-1` and `-1+y=n pi pm (pi//4), n in Z`
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