Solveeit Logo
Login

Question

Question: If the sides of triangle a, b, c be in A.P. then \(\tan \frac { A } { 2 } + \tan \frac { C } { 2 }\)...

If the sides of triangle a, b, c be in A.P. then tan⁡A2+tan⁡C2\tan \frac { A } { 2 } + \tan \frac { C } { 2 } equal to

A

23cot⁡A2\frac { 2 } { 3 } \cot \frac { A } { 2 }

B

23cot⁡B2\frac { 2 } { 3 } \cot \frac { B } { 2 }

C

23cot⁡C2\frac { 2 } { 3 } \cot \frac { C } { 2 }

D

None of these

Answer

None of these

Explanation

Solution

tan⁡A2+tan⁡C2=(s−b)(s−c)s(s−a)+(s−a)(s−b)s(s−c)\tan \frac { A } { 2 } + \tan \frac { C } { 2 } = \sqrt { \frac { ( s - b ) ( s - c ) } { s ( s - a ) } } + \sqrt { \frac { ( s - a ) ( s - b ) } { s ( s - c ) } }

= bscot⁡B2\frac { b } { s } \cot \frac { B } { 2 } = 2b2scot⁡B2\frac { 2 b } { 2 s } \cot \frac { B } { 2 }

∙∙\bullet \bullet a, b, c in A.P.

∴\therefore a+c=2ba + c = 2 b ⇒ 2s=3b2 s = 3 b = 2b3bcot⁡B2\frac { 2 b } { 3 b } \cot \frac { B } { 2 }

Hence, tan⁡A2+tan⁡C2=23cot⁡B2\tan \frac { A } { 2 } + \tan \frac { C } { 2 } = \frac { 2 } { 3 } \cot \frac { B } { 2 }.