Solveeit Logo
Login

Question

Mathematics Question on Trigonometric Functions

If A+B+C=π,A+B+C=\pi , then tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2+tan⁡C2tan⁡A2\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2} is equal to

A

33

B

22

C

11

D

00

Answer

11

Explanation

Solution

tan⁡A2.tan⁡B2+tan⁡B2.tan⁡C2+tan⁡C2.tan⁡A2\tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+\tan \frac{C}{2}.\tan \frac{A}{2}
(s−b) (s−c)s(s−a)  (s−a) (s−c)s (s−b)\sqrt{\frac{(s-b)\,(s-c)}{s(s-a)}}\,\,\sqrt{\frac{(s-a)\,(s-c)}{s\,(s-b)}}
+(s−a) (s−c)s(s−b)  (s−a) (s−b)s (s−a)+\sqrt{\frac{(s-a)\,(s-c)}{s(s-b)}}\,\,\sqrt{\frac{(s-a)\,(s-b)}{s\,(s-a)}}
=s−cs+s−as+s−bs=\frac{s-c}{s}+\frac{s-a}{s}+\frac{s-b}{s}
=3s−(a+b+c)s=3s−2ss=\frac{3s-(a+b+c)}{s}=\frac{3s-2s}{s}
(∵  2s=a+b+c)(\because \,\,2s=a+b+c)
=ss=1=\frac{s}{s}=1