If cos 2 B = c o s ( A + C ) c o s ( A − C ) , then tan A, tan B, tan C are in Options: A . &nbsp A.P. B . &nbsp G.P. C . &nbsp H.P. D . &nbsp None of these

Answer: Option B : B cos 2B = c o s ( A + C ) c o s ( A − C ) = c o s A c o s C − s i n A s i n C c o s A c o s C + s i n A s i n C ⇒ 1 − t a n 2 B 1 + t a n 2 B = 1 − t a n A t a n C 1 + t a n A t a n C ⇒ 1 + t a n 2 B − t a n A t a n C − t a n A t a n C t a n 2 B = 1 − t a n 2 B + t a n A t a n C − t a n A t a n C t a n 2 B ⇒ 2 t a n 2 B = 2 t a n A t a n C ⇒ t a n 2 B = tan A tan C Hence, tan A, tan B and tan C will be in G.P.

If cos 2 B = cos(A+C)cos(A−C), then tan A, tan B, tan C are in

Discussion Forum : Trigonometric Functions

Question -

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

Options:

A . A.P.

B . G.P.

C . H.P.

D . None of these

Answer: Option B
:
B

cos 2B =
$\frac{c o s (A + C)}{c o s (A - C)}$ =
$\frac{c o s A c o s C - s i n A s i n C}{c o s A c o s C + s i n A s i n C}$

$\Rightarrow$ $\frac{1 - t a n^{2} B}{1 + t a n^{2} B}$ =
$\frac{1 - t a n A t a n C}{1 + t a n A t a n C}$

$\Rightarrow$ $1 + t a n^{2} B - t a n A t a n C - t a n A t a n C t a n^{2} B$

= $1 - t a n^{2} B + t a n A t a n C - t a n A t a n C t a n^{2} B$

$\Rightarrow$ $2 t a n^{2} B = 2 t a n A t a n C$ $\Rightarrow$ $t a n^{2} B$ = tan A tan C

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

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