Q.If $\sin\theta = \dfrac{3}{5}$ and $\theta$ lies in the second quadrant, find $\cos\theta$ and $\tan\theta$.

Knowing $\sin\theta$ alone gives $\cos\theta$ only up to a sign through the Pythagorean identity; the quadrant given in the question is what fixes that sign correctly.

$\cos^2\theta = 1-\dfrac{9}{25}=\dfrac{16}{25}$, so $\cos\theta=\pm\dfrac{4}{5}$; in QII cosine is negative, so $\cos\theta=-\dfrac{4}{5}$, and $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=-\dfrac{3}{4}$.

Step 1 — apply the Pythagorean identity. $\sin^2\theta+\cos^2\theta=1 \Rightarrow \cos^2\theta = 1-\left(\dfrac{3}{5}\right)^2 = 1-\dfrac{9}{25}=\dfrac{16}{25}$, so $\cos\theta=\pm\dfrac{4}{5}$.

Step 2 — fix the sign using the quadrant. In the second quadrant ($90^\circ$ to $180^\circ$), by the All–Sin–Tan–Cos rule only sine (and cosecant) are positive; cosine is negative. So $\cos\theta=-\dfrac{4}{5}$.

Step 3 — find $\tan\theta$. $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{3/5}{-4/5}=-\dfrac{3}{4}$, consistent with tan also being negative in QII.

Check (independent route — verify both identities hold). $\sin^2\theta+\cos^2\theta = \dfrac{9}{25}+\dfrac{16}{25}=\dfrac{25}{25}=1$ ✓, and separately $1+\tan^2\theta=1+\dfrac{9}{16}=\dfrac{25}{16}$, while $\sec^2\theta = \left(\dfrac{1}{-4/5}\right)^2=\dfrac{25}{16}$ ✓ — both identities confirm the values.

Alternatively, picture a right triangle with opposite $=3$, hypotenuse $=5$, so adjacent $=\sqrt{25-9}=4$ by Pythagoras — then attach the correct quadrant sign to the adjacent side ($-4$) because the terminal side lies in QII, giving the same $\cos\theta=-4/5,\ \tan\theta=-3/4$.

Taking the positive root $\cos\theta=+4/5$ without checking the quadrant is the single most common error here; a second is getting the quadrant sign rule itself backwards (mixing up which ratios are positive in QII versus QIV).