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Question

Mathematics Question on Trigonometric Functions

If a=5,b=13,c=12a= 5, b = 13 , c = 12 in ΔABC\Delta ABC, then tan⁡B4\tan \frac{B}{4} is

A

3−1\sqrt{3} - 1

B

3+1\sqrt{3} + 1

C

2+1\sqrt{2} + 1

D

2−1\sqrt{2} - 1

Answer

2−1\sqrt{2} - 1

Explanation

Solution

tan⁡B4=sin⁡B4cos⁡B4\tan \frac{B}{4} = \frac{\sin \frac{B}{4}}{\cos \frac{B}{4}}
tan⁡B4=2sin⁡B4cos⁡B42cos⁡B4cos⁡B4\tan \frac{B}{4} = \frac{2\sin \frac{B}{4} \cos \frac{B}{4}}{2\cos \frac{B}{4}\cos \frac{B}{4}}
=sin⁡B21+cos⁡B2=sin⁡45∘1+cos⁡45∘       (∵  ∠B=90∘)= \frac{\sin \frac{B}{2}}{1+ \cos \frac{B}{2}} = \frac{\sin 45^{\circ}}{1+\cos45^{\circ} } \, \, \, \, \, \, \, (\because \, \, \angle B = 90^{\circ})
=121+12=12+1=2−1= \frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2} +1}=\sqrt{2} -1