In △PQR, right angled at Q,PQ=8 and PR=17. Find the six trigonometric ratios of the angle P
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Solution
Given that cosA=3537. Let us consider △ABC (see fig), ∠B=90∘, with AB=35 and AC=37d. By Pythagoras theorem, we have AC2=AB2+BC2 372=352+BC2 BC2=372−352 =1369−1225=144 ∴BC=√144=12 secA=ACAB=3735,tanA=BCAB=1235 Now, secA+tanA=3735+1235=4935,secA−tanA=3735−1235=2535 ∴secA+tanAsecA−tanA=49352535=4935×3525=4925