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If a secθ+b tanθ+c=0 and psecθ+qtanθ+r=θ, prove that (br−qc)2−(pc−ar)2=(ap−bp)2 |
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Answer» If a secθ+b tanθ+c=0 and psecθ+qtanθ+r=θ, prove that (br−qc)2−(pc−ar)2=(ap−bp)2 |
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