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If t a nalpha=x/(x+1) and t a nbeta=1/(2...

If `t a nalpha=x/(x+1)` and `t a nbeta=1/(2x+1)` , then `alpha+beta`
is equal to `pi//2` (b) `pi//3` (c) `pi//6` (d) `pi//4`

Text Solution

Verified by Experts

We know,
`tan(alpha+beta) =(tanalpha+tanbeta)/(1−tanalpha.tanbeta)`
Putting the values
`=>tan(alpha+beta)tan(alpha+beta)=(x(2x+1)+x+1)/((2x+1)(x+1)−x)`
`=>tan(alpha+beta)=(2x^2+2x+1)/(2x^2+2x+x+1−x)`
Solving further
`=>tan(alpha+beta)=(2x^2+2x+1)/(2x^2+2x+1)`
...
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