If `3tan(theta-15^0)=tan(theta+15^0),` then `theta` is equal to `n in Z)` n`pi+pi/4` (b) `npi+pi/8` `npi+pi/3` (d) none of these

Explanation

`3tan(theta-15^@) = tan(theta+15^@)`

`=>(tan(theta+15^@))/(tan(theta-15^@)) = 3/1`

Using componendo and dividendo,

`=>(tan(theta+15^@)+tan(theta-15^@))/(tan(theta+15^@)-tan(theta-15^@)) = (3+1)/(3-1)`

`=>(sin(theta+15^@)/cos(theta+15^@)+sin(theta-15^@)/cos(theta-15^@))/(sin(theta+15^@)/cos(theta+15^@)-sin(theta-15^@)/cos(theta-15^@)) = 2`

`=>(sin(theta+15^@)cos(theta-15^@)+sin(theta-15^@)cos(theta+15^@))/(sin(theta+15^@)cos(theta-15^@)-sin(theta-15^@)cos(theta+15^@)) = 2`

`=>(sin(theta+15^@+theta-15^@))/(sin(theta+15^@-theta+15^@)) = 2`

`=>(sin2theta)/(sin30^@) = 2`

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