1.

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

Answer» `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`
`=>(sin2theta)/(1/2) = 2`
`=>sin2theta = 1`
`=>2theta =2npi+ pi/2`
`=>theta = npi +pi/4 `
So, option-`(a)` is the correct option.


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