Question

In ∆PQR, ∠P = 90° and PQ = PR. The angle bisector of ∠QPR intersects QR at S, then prove that sin∠RPS/ sec∠PRS = sin2∠PQS.

Answer 1

Given :  In ∆PQR, ∠P = 90° and PQ = PR. The angle bisector of ∠QPR intersects QR at S,

To find : Prove that sin∠RPS  sec∠PRS = sin²∠PQS

Solution:

in  Δ PQR

PQ = PR

=> ∠Q = ∠R

∠P = 90°

∠P + ∠Q + ∠R = 180°

=> ∠Q + ∠R = 90°

=> ∠Q = ∠R  =  45°

PS is angle bisector   of 90°

=> ∠RPS = ∠QPS = 45°

sin∠RPS  = Sin45° = 1/√2

Sec ∠PRS = Sec∠R = Sec45°  = √2  

(sin∠RPS ) /(Sec  ∠PRS) = (1/√2)/√2   = 1/2

sin²∠PQS.  = Sin ²(∠Q) = Sin²(45°) = ( 1/√2)² = 1/2

1/2 = 1/2

(sin∠RPS ) /(Sec  ∠PRS) = sin²∠PQS.

QED

Hence Proved

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