Question

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

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is B

G.P.

cos 2B = $\frac{cos(A+C)}{cos(A-C)}$ = $\frac{cosAcosC-sinAsinC}{cosAcosC+sinAsinC}$

$\Rightarrow$$\frac{1-ta{n}^{2}B}{1+ta{n}^{2}B}$ = $\frac{1-tanAtanC}{1+tanAtanC}$

$\Rightarrow$$1+ta{n}^{2}B-tanAtanC-tanAtanCta{n}^{2}B$

= $1-ta{n}^{2}B+tanAtanC-tanAtanCta{n}^{2}B$

$\Rightarrow$ $2ta{n}^{2}B=2tanAtanC$ $\Rightarrow$ $ta{n}^{2}B$ = tan A tan C

Hence, tan A, tan B and tan C will be in G.P.