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    				| 1. | The general solution of the trigonometric equation `tan^2theta=1` isA. `theta=n pi pm(pi)/(3), n in z`B. `theta = npi pm(pi)/(6), n in z`C. `theta=n pi pm (pi)/(4), n in z`D. `theta = n pi, n in z` | 
| Answer» Correct Answer - C `tan^(2)theta = tan alpha` `theta= n pi pm alpha` `tan^(2)theta=1` `tan^(2) theta=tan^(2).(pi)/(4)` `therefore" "theta=n pi pm(pi)/(4)` | |