If $$\tan^2 \theta - 3 \sec \theta + 3 = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$ \sin \theta + \cot \theta$$ is:
We know
$$\tan^2\theta\ =\sec^2\theta\ -1$$
substituting we getÂ
$$\sec^2\theta\ -3\sec\theta\ +2\ =0$$
we get $$\sec\theta\ =2$$and $$\theta\ =60$$
so $$\sin\theta\ +\cot\theta\ =\frac{5\sqrt{\ 3}}{6}$$
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