Without tables or a calculator, evaluate exactly \(\displaystyle\frac{7\sin 33^\circ}{\cos 57^\circ}-\frac{2\,\mathrm{cosec}\,40^\circ}{\sec 50^\circ}\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5
In each fraction the two angles add up to \(90^\circ\). For complementary angles, \(\cos(90^\circ-A)=\sin A\) and \(\sec(90^\circ-A)=\mathrm{cosec}\,A\), so each fraction has the same value on the top and the bottom.
\(\cos 57^\circ=\cos(90^\circ-33^\circ)=\sin 33^\circ\), so \(\displaystyle\frac{7\sin 33^\circ}{\cos 57^\circ}=7\)
\(\sec 50^\circ=\sec(90^\circ-40^\circ)=\mathrm{cosec}\,40^\circ\), so \(\displaystyle\frac{2\,\mathrm{cosec}\,40^\circ}{\sec 50^\circ}=2\)
Value \(=7-2=5\)