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(a+2)sinalpha+(2a-1)cosalpha=(2a+1)iftan...

`(a+2)sinalpha+(2a-1)cosalpha=(2a+1)iftanalpha"i s"` `3/4` (b) `4/3` (c) `2a(a^2+1)` (d) `2a(a^2-1)`

A

`3//4`

B

`4//3`

C

`2a//(a^2+1)`

D

`2a//(a^2-1)`

Text Solution

Verified by Experts

The correct Answer is:
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)`
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