Q80 / 180 · COMEDK 2019 · Mathematics
The value of $\tan 67 \frac{1^{\circ}}{2}+\cot 67 \frac{1^{\circ}}{2}$ is
- A$\sqrt{2}$
- B$3 \sqrt{2}$
- Correct answerC$2 \sqrt{2}$
- D$2-\sqrt{2}$
Solution
We have,
$\tan 67 \frac{1}{2}^{\circ}+\cot 67 \frac{1}{2}^{\circ}$
$=\frac{\sin 67 \frac{1}{\circ}_{\circ}^{2}}{\cos 67 \frac{1}{2}^{\circ}}+\frac{\cos 67 \frac{1}{2}_{\circ}}{\sin 67 \frac{1}{2}^{\circ}}$
$=\frac{\sin ^{2} 67 \frac{1^{\circ}}{2}+\cos ^{2} 67 \frac{1}{2} \circ}{\cos 67 \frac{\frac{1}{2}^{\circ}}{2} \sin 67 \frac{1}{2} \circ}$
$=\frac{2}{2 \sin 67 \frac{1}{2}^{\circ} \cos 67 \frac{1}{2}^{\circ}}$
$=\frac{2}{\sin 135^{\circ}}$
$=\frac{2}{\sin \left(90^{\circ}+45^{\circ}\right)}$
$=\frac{2}{\cos 45^{\circ}}=2 \times \sqrt{2}=2 \sqrt{2}$
Hence, the correct answer is option: 'C'.