If $\tan^{-1}x+\tan^{-1}y=\frac{π}{4}$, then $\cot^{-1}x+\cot^{-1}y=$
Answer & explanation
Correct answer: option 2
$\frac{π}{2}-\cot^{-1}x+\frac{π}{2}-\cot^{-1}y=\frac{π}{4}$
$⇒\cot^{-1}x+\cot^{-1}y=\frac{3π}{4}$
If $\tan^{-1}x+\tan^{-1}y=\frac{π}{4}$, then $\cot^{-1}x+\cot^{-1}y=$
Correct answer: option 2
$\frac{π}{2}-\cot^{-1}x+\frac{π}{2}-\cot^{-1}y=\frac{π}{4}$
$⇒\cot^{-1}x+\cot^{-1}y=\frac{3π}{4}$