If A secθ + B tanθ + C = 0. and P secθ and Q tanθ + R = 0. Prove that: (BR – QC)2 – (PC – AR)2 = (AQ – BP)2
Formula: - (i) sec2θ – tan2θ = 1
Given: -
(i) Asecθ + Btanθ + C = 0
⇒ C = – (Asecθ + Btanθ)
(ii) Psecθ + Qtanθ + R = 0
⇒ R = – (Psecθ + Qtanθ)
taking L.H.S
(BR – QC)2 – (PC – AR)2 = ((B – (Psecθ + Qtanθ)) – (Q – (Asecθ + Btanθ)))2 – P( – (Asecθ + Btanθ)) – A( – (Psecθ + Qtanθ)))2
⇒ (BR – QC)2 – (PC – AR)2 = ( – BPsecθ – BQtanθ + AQsecθ + BQtanθ)2 – ( – APsecθ – BPtanθ + APsecθ + AQtanθ)2
⇒ (BR – QC)2 – (PC – AR)2 = (AQ – BP)2sec2θ – (AQ – BP)2tan2θ
⇒ (BR – QC)2 – (PC – AR)2 = (AQ – BP)2(sec2θ – tan2θ)
using formula(i)
⇒ (BR – QC)2 – (PC – AR)2 = (AQ – BP)2
L.H.S = R.H.S
Hence, Proved.
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