Question

If  PQ  and PR are equal sides of an isosceles right- angled triangle  PQR , then find the  value of  tan < P/2  +  cot < Q

Answer 1

u look at the diagram
PQ=PR
QR=√2PQ=√2PR
so <QPR=π/2
and <PQR+<PRQ=π/2
as PQ=PR
<PQR=<PRQ=π/4
so
$tan \frac{P}{2}=tan 22.5= \sqrt{2}-1, \\ cot Q=cot \frac{ \pi }{2}=0$

Answer 2

if u draw a diagram with given data,u notice that
PQ=PR which means angle R=angle Q
QR=PQ^1/2=PR^1/2
but given angle P=90 deg
hence,angle Q=angle R=90/2=45 deg
therefore,
tan<P/2+cot<Q=tan 90/2+cot 45
                       =tan 45+cot 45
                       =1+1=2....