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In triangle PQR PR=/x+3 , QR=/7 …

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In triangle PQR PR=/x+3 , QR=/7 and angle Q =90 evaluate Tan p.Sin R.Cosec P - Cos Q+ 7cot^ P. Cosec Q

    • Posted by Devyansh Joshi 45 6 years, 11 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Raj Dhali 6 years, 11 months ago

    PQ^2 = x-4 PQ=√x-4 evaluation=√7/√x-4 × √x-4/√x+3 × √x+3/√7 - 0(since cos90 is 0) =1-0=1

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