Question

If `(sinx)/(siny)=1/2,(cosx)/(cosy)=3/2` , where `x , y , in (0,pi/2),` then the value of `"tan"(x+y)` is equal to (a)`sqrt(13)` (b) `sqrt(14)` (c) `sqrt(17)` (d) `sqrt(15)`

Answer 1

`sinx/siny = 1/2 and cosx/cosy = 3/2`
`:. sinx/siny * cosy/cosx = (1/2)/(3/2)`
`=>tanx/tany = 1/3`
`=>tany = 3tanx`
`:. tan(x+y) = (tanx+tany)/(1-tanxtany) = (tanx+3tanx)/(1-tanx(3tanx))`
`=>tan(x+y) = (4tanx)/(1-3tan^2x)->(1)`
Now, `siny = 2sinx and cosy = 2/3cosx`
As, `sin^2y+cos^2y = 1`
`:. (2sinx)^2+(2/3cosx)^2 = 1`
`=>4sin^2x+4/9cos^2x = 1`
`=>4cos^2x(tan^2x + 1/9) = 1`
`=>4(tan^2x + 1/9)= sec^2x`
`=>4(tan^2x + 1/9)= 1+tan^2x`
`=>4tan^2x-tan^2x = 1-4/9`
`=>3tan^2x = 5/9`
`=>tan^2x = 5/27`
`=>tanx = sqrt5/(3sqrt3)`
Putting value of `tanx` in (1),
`tan(x+y) = (4* sqrt5/(3sqrt3))/(1-3(5/27)) = (4sqrt5sqrt3)/4 = sqrt15`
So option `(d)` is the correct option.