If `sinx+cos e cx+tany+coty=4` where `xa n dy in [0,pi/2],t h e ntany/2` is a root of the equation `alpha^2+2alpha+1=0` (b) 2`alpha^2-2alpha-1=0` `2alpha^2-2alpha-1=0` (d) `alpha^2-alpha-1=0`

Options

A. `a^(2)+2alpha+1=0`
B. `a^(2)+2a=0`
C. `2a^(2)-2a-1=0`
D. `a^(2)-alpha-1=0`

Answer

D

Explanation

`tan A-tanB=(1)/(2)`
`rArr(tanA)/(1)=(cotB)/(2)=gamma`(say)
`rArr tan A=gamma` and `tan B=(1)/(2lambda)`
`therefore (5-3cos 2A)(5-3cos2B)`
`=[5-(3(1-lambda^(2)))/((1+lambda^(2)))]xx[5-(3(1-(1)/(4lambda^(2))))/((1+(1)/(4lambda^(2))))]`