InterviewSolution
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InterviewSolution
| 1. |
Solve `4 cos theta-3` sec `theta=tan theta`. |
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Answer» We have `4 cos theta-3 sec theta = tan theta` or `4 cos theta-3/(cos theta)= (sin theta)/(cos theta)` or `4 cos^(2) theta-3= sin theta` or `4(1- sin^(2) theta)-3= sin theta` or `4 sin^(2) theta + sin theta-1=0` or `sin theta=(-1 pm sqrt(1+16))/8=(-1 pm sqrt(17))/8` Now, `sin theta=(-+sqrt(17))/8`. Thus, `sin theta=sin alpha`, where `sin alpha=(-1+sqrt(17))/8` `rArr theta=n pi +(-1)^(n) alpha`, where `sin alpha =(-1 + sqrt(17))/8 and n in Z` Also, `sin theta=(-1-sqrt(17))/8` `rArr sin theta= sin beta`, where `sin beta=(-1-sqrt(17))/8` `rArr theta=n pi +(-1)^(n) beta, n in Z` where `sin beta =(-1-sqrt(17))/8` |
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