Trigonometric Identities

The governing principle here is that a symmetric combination of sine and cosine, once squared, exposes their product through the Pythagorean identity, and the quadrant then disambiguates the individual signs. The key relation is $(\sin \theta +\cos \theta {)}^2={\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =1+2\sin \theta \cos \theta$, since the squared terms always collapse to one. Squaring the given value $\frac{1}{5}$ gives $\frac{1}{25}=1+2\sin \theta \cos \theta$, so $2\sin \theta \cos \theta =-\frac{24}{25}$, a negative product that already signals one positive and one negative ratio, consistent with the second quadrant. To separate the ratios, consider the difference squared: $(\sin \theta -\cos \theta {)}^2=1-2\sin \theta \cos \theta =1+\frac{24}{25}=\frac{49}{25}$, giving $\sin \theta -\cos \theta =\pm \frac{7}{5}$. In the second quadrant sine is positive while cosine is negative, so the difference must be positive, fixing the sign as $+\frac{7}{5}$. Solving the simultaneous pair $\sin \theta +\cos \theta =\frac{1}{5}$ and $\sin \theta -\cos \theta =\frac{7}{5}$ by adding and subtracting yields $\sin \theta =\frac{4}{5}$ and $\cos \theta =-\frac{3}{5}$, hence $\tan \theta =\frac{\sin \theta }{\cos \theta }=-\frac{4}{3}$. Option $-\frac{3}{4}$ inverts the ratio by swapping sine and cosine. Option $\frac{3}{4}$ ignores the quadrant sign that forces a negative tangent. Option $-\frac{4}{5}$ confuses the value of $\tan \theta$ with the magnitude of $\sin \theta$. This matches the standard JEE Advanced quadrant-and-sign analysis built on the fundamental identity. As a final plausibility check, ${\sin }^2\theta +{\cos }^2\theta =\frac{16}{25}+\frac{9}{25}=1$ holds exactly, confirming the recovered ratios are valid.