1.

General solution of `sin^2x-5sinxcosx-6cos^2x=0`is`x=npi-pi/4,n in Z`onlyn`pi+tan^(-1)6,n in Zon l y`both (a) and (b)none of these

Answer» `sin^2x/cos^2x-(5sinxcosx)/(cos^2x)-(6cos^2x)/(cos^2x)=0`
`tan^2x-5tanx-6=0`
`tan^2x-6tanx+tanx-6=0`
`tanx(tanx-6)+1(tanx-6)=0`
`(tanx+1)(tanx-6)=0`
`tanx=-1`
`x=npi-pi/4`
`tanx=6`
`v=npi+tan^(-1)6`
`tantheta=tanalpha`
`theta=npi+alpha`
Option C is correct.


Discussion

No Comment Found

Related InterviewSolutions