This question was uploaded on 03/10/21 on social media accounts.
Three angles (A, B, C) are angles of a triangle (Sum is 180). sin(A), sin(B), sin(C) are in A.P. Then find the value of tan(A/2).tan(C/2).
Solution:
sin(A), sin(B), sin(C) are in A.P:
\[\Rightarrow\sin(A)+\sin(C)=2\sin(B)\]
B = 180 - (A + C):
\[\Rightarrow\sin(A)+\sin(C)=2\sin(A+C)\]
\[\Rightarrow2\sin\left(\frac{A+C}{2}\right)\cdot\cos\left(\frac{A-C}{2}\right)=4\sin\left(\frac{A+C}{2}\right)\cdot\cos\left(\frac{A+C}{2}\right)\]
\[\Rightarrow\cos\left(\frac{A-C}{2}\right)=2\cos\left(\frac{A+C}{2}\right)\]
\[\Rightarrow\cos\left(\frac{A}{2}\right)\cos\left(\frac{C}{2}\right)+\sin\left(\frac{A}{2}\right)\sin\left(\frac{C}{2}\right)\]
\[=2\cos\left(\frac{A}{2}\right)\cos\left(\frac{C}{2}\right)-2\sin\left(\frac{A}{2}\right)\sin\left(\frac{C}{2}\right)\]
\[\Rightarrow 3\sin\left(\frac{A}{2}\right)\sin\left(\frac{C}{2}\right)=\cos\left(\frac{A}{2}\right)\cos\left(\frac{C}{2}\right)\]
\[\Rightarrow \tan\left(\frac{A}{2}\right)\tan\left(\frac{C}{2}\right)=\frac{1}{3}\]
Question related to Algebra, Equation, Trigonometry.
