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 $\tan x \tan y \tan z \tan t=-1$