KCET · Maths · Trigonometric Ratios & Identities
\(\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}\) is equal to
- A \(\frac{1}{\sqrt{3}}\)
- B \(\sqrt{3}\)
- C \(\frac{1}{2}\)
- D 1
Answer & Solution
Correct Answer
(B) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}} \\=& \frac{\sin \left(90^{\circ}-20^{\circ}\right)+\sin \left(90^{\circ}-40^{\circ}\right)}{\cos \left(90^{\circ}-20^{\circ}\right)+\sin \left(90^{\circ}-50^{\circ}\right)} \\=& \frac{\cos 20^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\cos 50^{\circ}} \\=& \frac{2 \cos 30^{\circ} \cos 10^{\circ}}{2 \cos 60^{\circ} \cos 10^{\circ}}=\frac{2 \cos 30^{\circ}}{2 \cos 60^{\circ}} \\=& \frac{2(\sqrt{3} / 2)}{2\left(\frac{1}{2}\right)}=\sqrt{3} \end{aligned}\)
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