0 votes 0 votes If $a,b,c$ are the sides of $\Delta ABC$, then $\tan \frac{B-C}{2} \tan \frac{A}{2}$ is equal to $\frac{b+c}{b-c}$ $\frac{b-c}{b+c}$ $\frac{c-b}{c+b}$ none of these Quantitative Aptitude isi2017-dcg quantitative-aptitude trigonometry geometry + – gatecse asked Sep 18, 2019 recategorized Nov 15, 2019 by Lakshman Bhaiya gatecse 257 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Using the Napier's Analogies(Laws of tangents) (Source: Law of tangents): $tan\left ( \frac{B-C}{2} \right )=\frac{b-c}{b+c}cot\left ( \frac{A}{2} \right )$ $\therefore$ $tan\left ( \frac{B-C}{2} \right )\times tan\left ( \frac{A}{2} \right )=\frac{b-c}{b+c}\times cot\left ( \frac{A}{2} \right )\times tan\left ( \frac{A}{2} \right )=\frac{b-c}{b+c}$ Option B is correct. haralk10 answered Mar 15, 2020 haralk10 comment Share Follow See all 0 reply Please log in or register to add a comment.