Question

If $\frac{{\sin }^2\mathrm{\theta }-3\mathrm{sin\theta }+2}{{\cos }^2\mathrm{\theta }}=1,\mathrm{then}\mathrm{\theta }=\mathrm{\_\_\_\_\_\_\_}.$

Answer 1

$$\begin{gathered} \frac{{\sin }^2\theta -3\sin \theta +2}{{\cos }^2\theta }=1 \\ \Rightarrow {\sin }^2\theta -3\sin \theta +2=1-{\sin }^2\theta \\ \Rightarrow 2{\sin }^2\theta -3\sin \theta +1=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2{\sin }^2\theta -2\sin \theta -\sin \theta +1=0 \\ \Rightarrow 2\sin \theta \left(\sin \theta -1\right)-1\left(\sin \theta -1\right)=0 \\ \Rightarrow \left(\sin \theta -1\right)\left(2\sin \theta -1\right)=0 \\ \Rightarrow \sin \theta -1=0\mathrm{or}2\sin \theta -1=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \sin \theta =1\mathrm{or}\sin \theta =\frac{1}{2} \\ \Rightarrow \theta =90°\mathrm{or}\theta =30° \end{gathered}$$
Here, $\theta$ cannot take the value 90º. For $\theta =90°$, the LHS of the given equation is not defined.

Thus, the value of $\theta$ is $30°$.

If $\frac{{\sin }^2\mathrm{\theta }-3\mathrm{sin\theta }+2}{{\cos }^2\mathrm{\theta }}=1,\mathrm{then}\mathrm{\theta }=\underline{30°}.$