If `angleA` and `angleP` are acute angle such that `tanA=tanP`, then show that `angleA=angleP`.

If A and B are acute angles such that tanA=cotB , then show that A+B =90^(@) .

If tanA=cotB and angles A and B are acute, then

If A, B and C are angles of a triangle such that angleA is obtuse, then show tan B tan C lt 1.

Giving reason show that anglex=anglea+angleb

If A and B are acute angles such that tanA=1/2 , tanB=1/3 and tan(A+B)=(tanA+tan\ B)/(1-tanAtanB) , find A+B .

In the adjoining figure, we have angle1=angle3 and angle2=angle4 . Show that angleA=angleC .

In an acute-angled triangle ABC, tanA+tanB+tanC

In figure, angleB lt angleA and angleC lt angleD . Show that AD lt BC .

If bisectors of angleA and angleB of a quadrilateral ABCD intersect each other at P, of angleB and angleC at Q, of angleC and angleD of R and of angleD and angleA at S, then PQRS is a (A) Rectangle (B) Rhombus (C) Parallelogram (D) Quadrilateral whose opposite angles are supplementary

Find angleP in the figure below.