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2 2Bk 9S00 pp‘ &% d =gue) + goos Sk |
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Answer» secA + tanA = p1/cosA + sinA/cosA = p(1 + sinA)/cosA = p(1 + sinA)^2/cosA^2 = p^2[squaring both sides](1 + sinA)^2/(1 - sinA^2) = p^2[ since cosA^2 + sinA^2 = 1](1 + sinA)^2/(1 + sinA)(1 - sinA) = p^2(1 + sinA)/(1 - sinA) = p^2[the common part getting cancelled]2/2sinA = (p^2 + 1)/(p^2 - 1)[By componendo and dividendo we get]1/sinA = (p^2 + 1)/(p^2 - 1) ThereforesinA = (p^2 - 1)/(p^2 + 1)cos A =√1 - sin^2 A =2p^2/(p^2 + 1) |
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