Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $a^2,b^2,c^2$ are in A.P., then which of the following is correct?

• A. $\frac{tan2B}{tan(A+C)}=\frac{1-tanA.tanC}{1-ta{n}^{2}B}$
• B. $\frac{tan(A+C)}{2tanB}=\frac{1-ta{n}^{2}B}{1-tanA.tanC}$
• C. $tanB.tan(A+C)=\frac{2tanA.tanC}{1-tanA.tanC}$
• D. $tanB.tan(A+C)=\frac{tanA.tanC}{1+2tanA.tanC}$

Answer 1

The correct option is C $tanB.tan(A+C)=\frac{2tanA.tanC}{1-tanA.tanC}$

$a^2,b^2,c^2$ are in A.P. $\Rightarrow$ $b^2-a^2=c^2-b^2$
$\Rightarrow$
${\sin }^{2}B-{\sin }^{2}A={\sin }^{2}C-{\sin }^{2}B$ or$\sin (B+A)\sin (B-A)=\sin (C+B)\sin (C-B)$ or $\sin C(\sin B\cos A-\cos BsinA)=\sin A(\sin C\cos B-\cos C\sin B)$
Dividing both sides by $\sin A\sin B\sin C,$ we get
$\cot A-\cot B=\cot B-\cot C$
$\Rightarrow$ $\cot A,\cot B,\cot C$are
in A.P.$\Rightarrow$ $\tan A,tanB,\tan C$ are in H.P
Hence, $\frac{2}{\tan B}=\frac{1}{\tan A}+\frac{1}{\tan C}$
$\Rightarrow \frac{2}{\tan B}=\frac{\tan A+\tan C}{\tan A.\tan C}=\frac{\tan (A+C)\times (1-\tan A\tan C)}{\tan A.\tan C}$
$\therefore \tan B.\tan (A+C)=\frac{2\tan A.\tan C}{1-\tan A.\tan C}$