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

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by debanti

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by debanti

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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

PQ=PR

QR=√2PQ=√2PR

so <QPR=π/2

and <PQR+<PRQ=π/2

as PQ=PR

<PQR=<PRQ=π/4

so

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....