Question

$3tan\theta$ + $cot\theta$ = 5 $cosec\theta$. Solve for $\theta$,0 $<\theta \le$ 90.

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Answer 1

$\frac{3sin\theta }{cos\theta }$ + $\frac{cos\theta }{sin\theta }$ = $\frac{5}{sin\theta }$

$sin\theta$ ( $3si{n}^2\theta$ + $co{s}^2\theta$) = 5 $sin\theta cos\theta$

3(1 - $co{s}^2\theta$) + $co{s}^2\theta$ = 5 $cos\theta$

2$co{s}^2\theta$ + 5$cos\theta$ - 3 = 0

2$cos\theta$ [$cos\theta$ + 3] - 1($cos\theta$ + 3) = 0

($cos\theta$ + 3) (2$cos\theta$ - 1) = 0

$cos\theta$ = -3 or $cos\theta$ = $\frac{1}{2}$

$cos\theta$ = -3 not possible.
Hence $\theta$ = ${60}^{\circ }$ from $cos\theta$ = $\frac{1}{2}$