Question

In triangle $ABC,\frac{\tan A}{2}=\frac{\tan b}{3}=\frac{\tan c}{4}$ then the value of ${\sec }^{2}A+{\sec }^{2}B+{\sec }^{2}C=$

• A. $\frac{101}{8}$
• B. $\frac{111}{8}$
• C. $\frac{121}{8}$
• D. $\frac{91}{8}$

Answer 1

Answer: (B)

In a $△ABC$, it is given that
$\frac{\tan A}{2}=\frac{\tan B}{3}=\frac{\tan C}{4}=k$
(Let) $\Rightarrow \tan A=2k,\tan B=3k$ and $\tan C=4k.$
Since in $△ABC$,
$\tan A+\tan B+\tan C=\tan A\tan B\tan C$
$\Rightarrow 9k=24k^3$
$\Rightarrow k^2=\frac{3}{8}\{\because k\ne 0\}$
$\Rightarrow 9k=24k^3$
$\Rightarrow k^2=\frac{3}{8}\{\because k\ne 0\}$
Now, ${\sec }^{2}A+{\sec }^{2}B+{\sec }^{2}C$
$=3+{\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C$
$=3+k^2[4+9+16]=3+\frac{3}{8}(29)=\frac{24+87}{8}=\frac{111}{8}$
$=3+{\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C$
$=3+k^2[4+9+16]=3+\frac{3}{8}(29)$
$=\frac{24+87}{8}=\frac{111}{8}$