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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