MHT CET · Maths · Trigonometric Ratios & Identities
If \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are the angles of a triangle with \(\tan \frac{A}{2}=\frac{1}{3}, \tan \frac{B}{2}=\frac{2}{3}\) then the value of \(\tan \frac{C}{2}\) is
- A \(\frac{-7}{9}\)
- B \(\frac{7}{9}\)
- C \(\frac{9}{7}\)
- D \(\frac{-9}{7}\)
Answer & Solution
Correct Answer
(B) \(\frac{7}{9}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & A+B+C=\pi \\ \therefore \quad & \tan \left(\frac{A+B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ \Rightarrow & \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\cot \frac{C}{2} \\ \Rightarrow & \frac{\frac{1}{3}+\frac{2}{3}}{1-\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)}=\cot \frac{C}{2}\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{9}{7}=\cot \frac{C}{2} \\ & \Rightarrow \tan \frac{C}{2}=\frac{7}{9}\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{9}{7}=\cot \frac{C}{2} \\ & \Rightarrow \tan \frac{C}{2}=\frac{7}{9}\end{aligned}\)
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