Question : If $\sin \theta-\cos \theta=0$, then what is the value of $\sin ^2 \theta+\tan ^2 \theta$ ?
Option 1: $\frac{1}{2}$
Option 2: $1$
Option 3: $\frac{4}{5}$
Option 4: $\frac{3}{2}$
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Correct Answer: $\frac{3}{2}$
Solution : Given $\sin \theta-\cos \theta=0$ ⇒ $\sin\theta = \cos\theta$ ⇒ $\frac{\sin\theta}{\cos\theta} = 1$ ⇒ $\tan\theta = 1$ ⇒ $\theta = 45°$ And $\sin\theta = \frac{1}{\sqrt{2}}$ [value of $\sin\theta$ at 45° is $\frac{1}{\sqrt{2}}$] Now, $\sin ^2 \theta+\tan ^2 \theta=(\frac{1}{\sqrt2})^2+1^2$ ⇒ $\frac{1}{2}+1 = \frac{3}{2}$ Hence, the correct answer is $\frac{3}{2}$.
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