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MHT CET · Maths · Trigonometric Ratios & Identities

If \(\cos 2 B-\frac{\cos (A+C)}{\cos (A-C)}\), Then \(\tan A, \tan B, \tan C\) are in

  1. A Geometric Progression.
  2. B Arithmetic Progression.
  3. C Harmonic Progression.
  4. D Arithmetico-Geometric Progression.
Verified Solution

Answer & Solution

Correct Answer

(A) Geometric Progression.

Step-by-step Solution

Detailed explanation

\(\cos 2 B=\frac{\cos (A+C)}{\cos (A-C)} \)
\( \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C} \)
\( \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{\cos A \cos C(1-\tan A \tan C)}{\cos A \cos C(1+\tan A \tan C)} \)
\( \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{(1-\tan A \tan C)}{(1+\tan A \tan C)}\)
\(\left(1-\tan ^2 \mathrm{~B}\right)(1+\tan \mathrm{A} \tan \mathrm{C}) \)
\( =(1-\tan \mathrm{A} \tan \mathrm{C})\left(1+\tan ^2 \mathrm{~B}\right) \)
\( 1+\tan \mathrm{A} \tan \mathrm{C}-\tan ^2 \mathrm{~B}-\tan ^2 \mathrm{~B} \tan \mathrm{A} \tan \mathrm{C} \)
\( =1+\tan ^2 \mathrm{~B}-\tan \mathrm{A} \tan \mathrm{C}-\tan \mathrm{A} \tan \mathrm{C} \tan ^2 \mathrm{~B} \)
\( 2 \tan \mathrm{A} \tan \mathrm{C}=2 \tan ^2 \mathrm{~B} \)
\( \tan ^2 \mathrm{~B}=\tan \mathrm{A} \cdot \tan \mathrm{C} \)
\( \therefore \tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C} \text { are in G.P. }\)