Question:

<div class="solution"><p><strong>Step 1:</strong> <span class="math-inline">\(\cos^{-1}(1/\sqrt{5})=\theta\implies\sin\theta=2/\sqrt{5}\)</span>. <span class="math-inline">\(\sin 2\theta=2\cdot(2/\sqrt{5})\cdot(1/\sqrt{5})=4/5\)</span>.</p><p><strong>Step 2:</strong> <span class="math-inline">\(\tan^{-1}(1/3)=\alpha\)</span>. <span class="math-inline">\(\cos 2\alpha=\frac{1-1/9}{1+1/9}=4/5\)</span>.</p><p><strong>Step 3:</strong> Sum <span class="math-inline">\(=4/5+4/5=8/5\)</span>. So <span class="math-inline">\(p=8,q=5,p-q=3\)</span>.</p><p><strong>Step 4:</strong> Units digit of <span class="math-inline">\(3^{2k+1}\)</span>: odd powers of 3 cycle 3,7,3,7,... → units digits 3 or 7.</p><p><strong>Answer: (B),(C)</strong></p><div class="trap-box"><strong>Trap:</strong> Odd powers of 3 alternate between units digits 3 and 7.</div><div class="key-concept"><strong>Key Concept:</strong> Double-angle formulas + cyclicity of units digits of powers</div></div>