0606 P12 - Jun 2025 - Q4 - 5 marks
7118
(a) Find \(\int_0^\pi \sin\theta\,\mathrm{d}\theta\).
(b) Given that \(0\lt \alpha\lt \frac{\pi}{2}\), show that \(\frac{\sec\alpha}{\cot\alpha+\tan\alpha}\) can be written as \(\sin\alpha\).
Solution
Answer: (a) \(2\). (b) \(\frac{\sec\alpha}{\cot\alpha+\tan\alpha}=\sin\alpha\).
(a) Since \(\int\sin\theta\,\mathrm{d}\theta=-\cos\theta\),
\(\int_0^\pi \sin\theta\,\mathrm{d}\theta=[-\cos\theta]_0^\pi.\)
Therefore
\(-\cos\pi-(-\cos0)=1+1=2.\)
(b) Write all three ratios in terms of sine and cosine:
\(\frac{\sec\alpha}{\cot\alpha+\tan\alpha} =\frac{\frac{1}{\cos\alpha}}{\frac{\cos\alpha}{\sin\alpha}+\frac{\sin\alpha}{\cos\alpha}}.\)
The denominator is
\(\frac{\cos^2\alpha+\sin^2\alpha}{\sin\alpha\cos\alpha} =\frac{1}{\sin\alpha\cos\alpha}.\)
So
\(\frac{\frac{1}{\cos\alpha}}{\frac{1}{\sin\alpha\cos\alpha}} =\frac{1}{\cos\alpha}\cdot\sin\alpha\cos\alpha =\sin\alpha.\)
