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If `tan theta = p/q` and `theta = 3 phi( 0 < theta < pi/2)`, then `p/(sin phi) - q/ (cos phi)` = |
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Answer» Here, ` tantheta = p/q=>p = qtantheta->(1)` `=>sec theta = sqrt(1+p^2/q^2) = sqrt(p^2+q^2)/q` `=>costheta = q/(sqrt(p^2+q^2))->(2)` `:. p/sinphi - q/cosphi = (qtantheta)/sinphi - q/cosphi` `=q(tantheta/sinphi - 1/cosphi)` `=q(sintheta/(costheta sinphi) - 1/cosphi)` `=q((sinthetacosphi -costhetasinphi)/(costheta sinphicosphi))` `=2q(sin(theta-phi)/(2sinphicosphi(costheta)))` `=2q(sin(3phi-phi)/(sin2phi(costheta)))` .....(As `theta = 3phi`) `=(2q)/costheta` `=2q(sqrt(p^2+q^2)/q)` .....From(2) `=2sqrt(p^2+q^2)` `:. p/sinphi - q/cosphi = 2sqrt(p^2+q^2)` |
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