The angle \(A\) is obtuse and \(\sin A = \dfrac{3}{5}\). The angle \(B\) is acute and \(\cos B = \dfrac{5}{13}\). Show that \(\cos A = -\dfrac{4}{5}\), and ...

This question is about the addition formulae, but the marks turn on something more elementary: choosing the correct sign when a Pythagorean identity is used to recover a second ratio. The words "obtuse" and "acute" are the whole reason the question is set this way, and ignoring them is what the question is built to catch.

(a) Showing \(\cos A = -\dfrac{4}{5}\) and finding \(\sin B\) [2 marks]

Rearranging \(\sin^{2} A + \cos^{2} A \equiv 1\) gives

\[\cos^{2} A = 1 - \sin^{2} A = 1 - \frac{9}{25} = \frac{16}{25}, \qquad \text{so } \cos A = \pm\frac{4}{5}.\]

The square root produces two candidates, and the context decides between them. An obtuse angle lies between \(90^{\circ}\) and \(180^{\circ}\), which is the second quadrant, where the cosine is negative. Hence

\[\cos A = -\frac{4}{5} \qquad \textbf{B1}\]

Because the target is printed in the question, the mark is for the reasoning, not the statement: a candidate who writes \(\cos A = -\tfrac{4}{5}\) with no identity and no appeal to the quadrant earns nothing. The same identity applied to \(B\), which is acute and therefore in the first quadrant where every ratio is positive, gives

\[\sin^{2} B = 1 - \frac{25}{169} = \frac{144}{169}, \qquad \sin B = +\frac{12}{13} \qquad \textbf{B1}\]

The two B marks are independent, so a sign error on \(\cos A\) does not cost the \(\sin B\) mark. Both results come from the triples \(3, 4, 5\) and \(5, 12, 13\), so untidy surds at this stage signal a slip.

(b) The exact value of \(\sin\left(A + B\right)\) [3 marks]

Use the addition formula for sine:

\[\sin\left(A + B\right) = \sin A\cos B + \cos A\sin B \qquad \textbf{M1}\]

The M1 is for quoting the correct expansion, and it stands even if the substituted values are wrong, which is why writing the formula down before substituting is worth doing. Substituting all four ratios, including the negative sign on \(\cos A\):

\[\sin\left(A + B\right) = \frac{3}{5} \times \frac{5}{13} + \left(-\frac{4}{5}\right) \times \frac{12}{13} = \frac{15}{65} - \frac{48}{65} \qquad \textbf{M1}\] \[= -\frac{33}{65} \qquad \textbf{A1}\]

Since \(33\) and \(65\) share no factor, the fraction is already in lowest terms. A negative answer is exactly what should be expected: \(A\) is obtuse and \(B\) is acute, so \(A + B\) can exceed \(180^{\circ}\) and place the sum in the third quadrant, where the sine is negative. A candidate who takes \(\cos A = +\tfrac{4}{5}\) obtains \(+\dfrac{63}{65}\), which loses both the second method mark and the accuracy mark. The other classic error is confusing the formulae and writing \(\sin A\cos B - \cos A\sin B\), which is the expansion of \(\sin\left(A - B\right)\).

(c) The exact value of \(\tan\left(A - B\right)\) [3 marks]

The tangent addition formula needs the two tangents, obtained as the ratio of sine to cosine:

\[\tan A = \frac{3/5}{-4/5} = -\frac{3}{4}, \qquad \tan B = \frac{12/13}{5/13} = \frac{12}{5} \qquad \textbf{B1}\]

Both come out with the fifths and thirteenths cancelling, which is why building the tangent from the two ratios is safer than reaching for a triangle. Note the sign: \(\tan A\) is negative because \(A\) is obtuse. Now apply the formula:

\[\tan\left(A - B\right) = \frac{\tan A - \tan B}{1 + \tan A\tan B} = \frac{-\dfrac{3}{4} - \dfrac{12}{5}}{1 + \left(-\dfrac{3}{4}\right)\left(\dfrac{12}{5}\right)} \qquad \textbf{M1}\]

Work out numerator and denominator over a common denominator of \(20\):

\[\text{numerator} = -\frac{15}{20} - \frac{48}{20} = -\frac{63}{20}, \qquad \text{denominator} = 1 - \frac{36}{20} = -\frac{16}{20}\] \[\tan\left(A - B\right) = \frac{-63/20}{-16/20} = \frac{63}{16} \qquad \textbf{A1}\]

Because \(63\) and \(16\) are coprime, this is in lowest terms. Two sign traps sit in this part. The first is the sign pattern of the formula itself: for \(A - B\) the numerator subtracts and the denominator adds, the opposite arrangement from \(A + B\). The second is the double negative in the denominator, where \(1 + \left(-\tfrac{3}{4}\right)\left(\tfrac{12}{5}\right)\) is \(1 - \tfrac{9}{5}\), a negative quantity; two negatives divided give the positive answer \(\dfrac{63}{16}\). Check the size against the geometry: \(A \approx 143.1^{\circ}\) and \(B \approx 67.4^{\circ}\), so \(A - B \approx 75.8^{\circ}\), whose tangent is about \(3.94\), and \(\dfrac{63}{16} = 3.9375\).

Examination takeaway. Whenever an identity forces a square root, write \(\pm\) explicitly and then quote the quadrant that fixes the sign. Learn the addition formulae as a family and check the sign pattern before substituting, because in questions of this type almost every lost mark is a lost sign rather than lost arithmetic.