In triangle ABC, right-angled at B, if tan A = 1 , 3 find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C
Step-by-Step Solution
Key Concept: Use the complementary relationship in a right‑angled triangle (A + C = 90°) and the sum‑to‑product identities: \(\sin(A+C)=\sin A\cos C+\cos A\sin C\) and \(\cos(A+C)=\cos A\cos C-\sin A\sin C\). The given \(\tan A\) helps to compute \(\sin A\) and \(\cos A\) if required.
1. Since \(\triangle ABC\) is right‑angled at \(B\), we have
$$A + C = 90^{\circ}.$$
2. From the given \(\tan A = \frac{1}{3}\), consider a right‑angled triangle with opposite side = 1 and adjacent side = 3.
\[\text{hypotenuse}=\sqrt{1^{2}+3^{2}}=\sqrt{10}.\]
Hence
$$\sin A = \frac{1}{\sqrt{10}}, \qquad \cos A = \frac{3}{\sqrt{10}}.$$
3. Because \(C = 90^{\circ} - A\), the complementary‑angle relations give
$$\sin C = \cos A = \frac{3}{\sqrt{10}}, \qquad \cos C = \sin A = \frac{1}{\sqrt{10}}.$$
4. Part (i)
\[\sin A\cos C + \cos A\sin C = \sin A\cdot\sin A + \cos A\cdot\cos A
= \left(\frac{1}{\sqrt{10}}\right)^{2}+\left(\frac{3}{\sqrt{10}}\right)^{2}
= \frac{1}{10}+\frac{9}{10}=1.\]
Alternatively, using the sum formula:
$$\sin(A+C)=\sin 90^{\circ}=1.$$
5. Part (ii)
\[\cos A\cos C - \sin A\sin C = \cos A\cdot\sin A - \sin A\cdot\cos A
= \frac{3}{\sqrt{10}}\cdot\frac{1}{\sqrt{10}}-\frac{1}{\sqrt{10}}\cdot\frac{3}{\sqrt{10}}=0.\]
Or, using the cosine sum formula:
$$\cos(A+C)=\cos 90^{\circ}=0.$$
6. Hence the required values are:
\[\text{(i)}\;=1, \qquad \text{(ii)}\;=0.\]
Correct Answer: (i) 1, (ii) 0