Search any question & find its solution
Question:
Answered & Verified by Expert
In triangle \(A B C, \frac{\tan A}{2}=\frac{\tan B}{3}=\frac{\tan C}{4}\), then the value of \(\sec ^2 A+\sec ^2 B+\sec ^2 C=\)
Options:
Solution:
2480 Upvotes
Verified Answer
The correct answer is:
\(\frac{111}{8}\)
In a \(\triangle A B C\), it is given that
\(\frac{\tan A}{2}=\frac{\tan B}{3}=\frac{\tan C}{4}=k \text { (Let) }\)
\(\Rightarrow \quad \tan A=2 k, \tan B=3 k\) and \(\tan c=4 k\).
Since in \(\triangle A B C\),
\(\begin{aligned}
& \tan A+\tan B+\tan C=\tan A \tan B \tan C \\
& \Rightarrow \quad 9 k=24 k^3 \Rightarrow k^2=\frac{3}{8} \quad\{\because k \neq 0\} \\
& \text {Now, } \sec ^2 A+\sec ^2 B+\sec ^2 C \\
& =3+\tan ^2 A+\tan ^2 B+\tan ^2 C \\
& =3+k^2[4+9+16]=3+\frac{3}{8}(29)=\frac{24+87}{8}=\frac{111}{8}.
\end{aligned}\)
\(\frac{\tan A}{2}=\frac{\tan B}{3}=\frac{\tan C}{4}=k \text { (Let) }\)
\(\Rightarrow \quad \tan A=2 k, \tan B=3 k\) and \(\tan c=4 k\).
Since in \(\triangle A B C\),
\(\begin{aligned}
& \tan A+\tan B+\tan C=\tan A \tan B \tan C \\
& \Rightarrow \quad 9 k=24 k^3 \Rightarrow k^2=\frac{3}{8} \quad\{\because k \neq 0\} \\
& \text {Now, } \sec ^2 A+\sec ^2 B+\sec ^2 C \\
& =3+\tan ^2 A+\tan ^2 B+\tan ^2 C \\
& =3+k^2[4+9+16]=3+\frac{3}{8}(29)=\frac{24+87}{8}=\frac{111}{8}.
\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.