Find the value of \(\displaystyle\frac{\sin70^\circ}{\cos20^\circ}+\frac{\tan46^\circ}{\cot44^\circ}+\cos49^\circ\cdot\text{cosec}\,41^\circ\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3
In each pair the two angles add up to \(90^\circ\), so complementary-angle identities apply: \(\sin\theta=\cos(90^\circ-\theta)\), \(\tan\theta=\cot(90^\circ-\theta)\) and \(\cos\theta=\sin(90^\circ-\theta)\).
\(\sin70^\circ=\cos20^\circ\), so \(\displaystyle\frac{\sin70^\circ}{\cos20^\circ}=1\)
\(\tan46^\circ=\cot44^\circ\), so \(\displaystyle\frac{\tan46^\circ}{\cot44^\circ}=1\)
\(\cos49^\circ=\sin41^\circ\), so \(\displaystyle\cos49^\circ\cdot\text{cosec}\,41^\circ=\frac{\sin41^\circ}{\sin41^\circ}=1\)
The three terms add up to \(1+1+1=3\)