निम्नलिखित समीकरणों से `theta` के व्यापक मान कीजिए :
`3 tan (theta- 15^(@)) = tan (theta + 15^(@))`

Explanation

` 3 tan (theta - 15^(@)) = tan (theta + 15^(@))`
` rArr (3 sin (theta - 15^(@)))/(cos (theta - 15^(@)) )= (sin (theta+15^(@)))/(cos (theta+15^(@)))`
` rArr 3 cos ( theta + 15^(@) )sin (theta - 15^(@))`
` = sin (theta + 15^(@) )cos (theta- 15^(@))`
` rArr 3 * {2 cos ( theta +15^(@))sin (theta- 15^(@))}`
` 2 sin ( theta + 15^(@)) * cos (theta- 15^(@))`
` rArr 3 ( sin 2 theta - sin 30^(@)) = sin 2 theta+ sin 30^(@)`
` rArr 2 sin 2 theta = 4 sin 30^(@) = 4 xx 1/2 `
` rArr sin 2theta = 1 = sin. pi/ 2`
` :. 2 theta = 1 = sin/pi/2 `
` :. 2 theta = n pi + (-1) ^(n) pi/2 `
` rArr theta = (npi)/2 + (-1)^(n) pi/2 `