Q.If $\tan\theta = 3$ and $\theta$ lies in third quadrant, then the value of $\sin\theta$ is
(A) $\dfrac{1}{\sqrt{10}}$
(B) $-\dfrac{1}{\sqrt{10}}$
(C) $\dfrac{-3}{\sqrt{10}}$
(D) $\dfrac{3}{\sqrt{10}}$

Concept: Trigonometric ratios from tangent using the Pythagorean identity, with sign determined by quadrant.

Given $\tan\theta = 3$, we know $\tan\theta = \frac{\sin\theta}{\cos\theta} = 3$, so $\sin\theta = 3\cos\theta$.

Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:

In the third quadrant, both sine and cosine are negative. Therefore $\cos\theta = -\frac{1}{\sqrt{10}}$. …

The key to this problem is understanding how trigonometric functions behave in different quadrants. The tangent of an angle is the ratio $\frac{\sin\theta}{\cos\theta}$, and its sign depends on the signs of sine and cosine. In the third quadrant (where angles lie between $180°$ and $270°$, or $\pi$ and $\frac{3\pi}{2}$ radians), both $x$ and $y$ coordinates are negative, which means both $\sin\theta$ and $\cos\theta$ are negative. Since tangent is the ratio of two negative quantities, it becomes positive—which is consistent with our given $\tan\theta = 3 > 0$.

Now we need to find the actual value of $\sin\theta$.

Step-by-step solution:

1. Start with the fundamental identity relating tangent and secant.

   We know that $1 + \tan^2\theta = \sec^2\theta$. Substituting $\tan\theta = 3$:

2. Find $\cos\theta$ from $\sec\theta$.

   Since $\sec\theta = \frac{1}{\cos\theta}$, we have:

Because $\theta$ is in the third quadrant where cosine is negative:

3. Use the tangent definition to find $\sin\theta$.

   We know $\tan\theta = \frac{\sin\theta}{\cos\theta} = 3$, so:

Substituting our value for $\cos\theta$:

4. Verify the quadrant condition. …