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If (sinx)/(siny)=1/2,(cosx)/(cosy)=3/2 ...

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)`

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`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`
...
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