Q.If $A$ lies in the second quadrant and $3\tan A + 4 = 0$, then the value of $2\cot A - 5\cos A + \sin A$ is equal to
(A) $\dfrac{-53}{10}$
(B) $\dfrac{23}{10}$
(C) $\dfrac{37}{10}$
(D) $\dfrac{7}{10}$

Concept: Trigonometric ratios in the second quadrant, where sine is positive while cosine and tangent are negative.

From $3\tan A + 4 = 0$, we get $\tan A = -\frac{4}{3}$.

Since $\tan A = \frac{\sin A}{\cos A} = -\frac{4}{3}$ and $A$ is in the second quadrant, we can represent this on a reference triangle: opposite side $= 4$, adjacent side $= -3$ (negative because cosine is negative in Q2). The hypotenuse is $\sqrt{4^2 + 3^2} = 5$.

Therefore:

• $\sin A = \frac{4}{5}$ (positive in Q2)
• $\cos A = -\frac{3}{5}$ (negative in Q2)
• $\cot A = \frac{1}{\tan A} = -\frac{3}{4}$

Substituting into the expression: …

The heart of this problem is understanding how the signs of trigonometric functions change across quadrants. In the second quadrant, sine is positive (the $y$-coordinate) while cosine is negative (the $x$-coordinate), making tangent negative. Once we know $\tan A$, we can reconstruct the exact values of $\sin A$ and $\cos A$ using the fundamental identity $\sin^2 A + \cos^2 A = 1$, being careful to pick signs that match the quadrant.

Let me work through this systematically.

1. Find $\tan A$ from the given equation

We have $3\tan A + 4 = 0$, so:

2. Interpret $\tan A$ geometrically

Recall that $\tan A = \frac{\sin A}{\cos A} = \frac{\text{opposite}}{\text{adjacent}}$ in a right triangle. If we think of a reference triangle with opposite side $4$ and adjacent side $3$, the hypotenuse by Pythagoras is:

3. Determine the signs in the second quadrant

In quadrant II:

• $\sin A > 0$ (positive $y$-coordinate)
• $\cos A < 0$ (negative $x$-coordinate)
• $\tan A < 0$ (negative/positive = negative) ✓

So with our reference triangle:

$\sin A = \frac{4}{5}, \quad \cos A = -\frac{3}{5}$ …