Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If $\cos 2 B-\frac{\cos (A+C)}{\cos (A-C)}$, Then $\tan A, \tan B, \tan C$ are in
MathematicsTrigonometric Ratios & IdentitiesMHT CETMHT CET 2023 (09 May Shift 2)
Options:
  • A Geometric Progression.
  • B Arithmetic Progression.
  • C Harmonic Progression.
  • D Arithmetico-Geometric Progression.
Solution:
1586 Upvotes Verified Answer
The correct answer is: Geometric Progression.
$\begin{aligned} & \cos 2 B=\frac{\cos (A+C)}{\cos (A-C)} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{\cos A \cos C(1-\tan A \tan C)}{\cos A \cos C(1+\tan A \tan C)} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{(1-\tan A \tan C)}{(1+\tan A \tan C)}\end{aligned}$
$\begin{aligned} &\left(1-\tan ^2 \mathrm{~B}\right)(1+\tan \mathrm{A} \tan \mathrm{C}) \\ & \quad=(1-\tan \mathrm{A} \tan \mathrm{C})\left(1+\tan ^2 \mathrm{~B}\right) \\ & \quad 1+\tan \mathrm{A} \tan \mathrm{C}-\tan ^2 \mathrm{~B}-\tan ^2 \mathrm{~B} \tan \mathrm{A} \tan \mathrm{C} \\ & \quad=1+\tan ^2 \mathrm{~B}-\tan \mathrm{A} \tan \mathrm{C}-\tan \mathrm{A} \tan \mathrm{C} \tan ^2 \mathrm{~B} \\ & \quad 2 \tan \mathrm{A} \tan \mathrm{C}=2 \tan ^2 \mathrm{~B} \\ & \quad \tan ^2 \mathrm{~B}=\tan \mathrm{A} \cdot \tan \mathrm{C} \\ & \therefore \quad \tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C} \text { are in G.P. }\end{aligned}$

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.