Question

If $\cos \left(2{\theta }_2\right)=\frac{\cos \left({\theta }_1+{\theta }_3\right)}{\cos \left({\theta }_1-{\theta }_3\right)}$, then

• A. $\tan {\theta }_1,\tan {\theta }_2,\tan {\theta }_3$ are in A.P.
• B. $\tan {\theta }_1,\tan {\theta }_2,\tan {\theta }_3$ are in G.P.
• C. $\tan {\theta }_1,\tan {\theta }_2,\tan {\theta }_3$ are in H.P.
• D. $\tan {\theta }_1,\tan {\theta }_2,\tan {\theta }_3$ are NOT in A.P./G.P./H.P.

Answer 1

Answer: (B)

$\cos \left(2{\theta }_2\right)=\frac{\cos \left({\theta }_1+{\theta }_3\right)}{\cos \left({\theta }_1-{\theta }_3\right)}\Rightarrow \frac{\cos \left(2{\theta }_2\right)}{1}=\frac{\cos \left({\theta }_1+{\theta }_3\right)}{\cos \left({\theta }_1-{\theta }_3\right)}$ (Using dividendo and componendo)
$\Rightarrow \frac{\cos \left(2{\theta }_2\right)-1}{\cos \left(2{\theta }_2\right)+1}=\frac{-2\sin {\theta }_1\sin {\theta }_3}{2\cos {\theta }_1\cos {\theta }_3}$
$\Rightarrow {\tan }^2{\theta }_2=\tan {\theta }_1\tan {\theta }_3\Rightarrow \tan {\theta }_1,\tan {\theta }_2,\tan {\theta }_3\text{ are in G.P.Ans.}$