In `DeltaABC,BC=1,sin.(A)/2=x_1,sin.(B)/2=x_2,cos.(A)/2=x_3andcos.(B)/2=x_4" with "(x_1/x_2)^2007-(x_3/x_4)^2006=0`
If `angleA=90^@`, then area of `DeltaABC` is

Options

A. `1/2`
B. 1
C. 2
D. can't be determined

Answer

B

Explanation

In given `DeltaABC" both " A/2and B/2"lie strictly in "(0,pi/2)and sin x " always increasing in "(0,pi/2)` whereas cos x is always decreasing in `(0,pi/2)`.
So, if `A/2ltB/2`
`rArr sin(A)/2gt sin(B)/2`
`or x_1gtx_2`
`and x_3ltx_4`
`or 1/x^3gt1/x^4`
So, `x_1^2007x_4^2006=x_2^2007x_3^2006` is not valid.
Similarly for `A/2ltB/2`
`rArr sin(A)/2sin(B)/2`
`rArr x_1ltx_2`
`and 1/x^3lt1/x^4`
For this also `x_1^2007x_4^2006=x_2^2007x_3^2006` is not valid.
So,`(x_1/x_2)^2007-(x_3/x_4)^2006=0" is possible only when " A/2=B/2`.
`rArr x_1=x_2and 1/x_3=1/x_4`
Hence, `DeltaABC` is isosceles with `angleABC=angleCAB`.
`rArr BC=AC=1"unit"`
if `angleA=90^@`
Area, `A=1/2BCxxAC=1/2" sq. units"`