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If tantheta=1/2a n dt a nvarphi=1/3, th...

If `tantheta=1/2a n dt a nvarphi=1/3,` then the value of `theta+varphi` is `pi/6` (b) `pi` (c) 0 (d) `pi/4`

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Given,
`tantheta=1/2 and tanvarphi=1/3`
we know that,
`=>tan(theta+phi)=(tanphi+tantheta)/(1-tanphitantheta)`
`=>tan(theta+phi)=((1/2)+(1/3))/(1-(1/2)(1/3))`
`=>tan(theta+phi)=(5/6)/((5/6)`
`=>tan(theta+phi)=1`
`=>(theta+phi)=tan^(-1)(1)`
...
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