Question

If $\frac{\cos (x-y)}{\cos (x+y)}+\frac{\cos (z+t)}{\cos (z-t)}=0$, then the value of $\tan x\tan y$ $\tan z\tan t$ is equal to

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (B)

$\frac{\cos (x-y)}{\cos (x+y)}+\frac{\cos (z+t)}{\cos (z-t)}=0$
or $\frac{1+\tan x\tan y}{1-\tan x\tan y}+\frac{1-\tan z\tan t}{1+\tan z\tan t}=0$
or $1+\tan z\tan t+\tan x\tan y$
$+\tan x\tan y\tan z\tan t+1-\tan z\tan t$
$-\tan x\tan y+\tan x\tan y\tan z\tan t=0$
or $\tan x\tan y\tan z\tan t=-1$