1.

`(a+2)sinalpha(2a-1)cosalpha=(2a+1)if tanalpha` isA. `3//4`B. `4//3`C. `2a//(a^2+1)`D. `2a//(a^2-1)`

Answer» Correct Answer - B::D
Divide by `cosalpha` and square both sides abd let `tanalpha=t` so that ` sec^2alpha=1+t^2`
`rArr [(a+2)t+(2a-1)]^2=[(2a+1)^2(1+t^2)]`
`rArr t^3[(a+2)^2-(2a+1)^2]+2(a+2)(2a-1)t+[(2a-1)^2-(2a+1)^2]=0`
`or 3(1-a^2)t^2+2(2a^2+3a-2)t-4xx2a=0`
`or 3(1-a^2)t^2+2(2a^2+3a-2)t-4xx2a=0`
`or t(1-a^2)t^2-4(1-a^2)t-6at-8a=0`
`or (3t-4)[(1-a^2)t+2a]=0`
`or t-tanalpha=4/3or(2a)/(a^2-1)`


Discussion

No Comment Found

Related InterviewSolutions