Resolve u into ux = u cos(theta) and uy = u sin(theta). Vertical motion has acceleration -g. Time of flight: uy - g(T/2) = 0 at the top, so T = 2u sin(theta)/g. Maximum height: H = uy^2/2g = u^2 sin^2(theta)/2g. Horizontal range: R = ux * T = u cos(theta) * 2u sin(theta)/g = u^2 sin(2 theta)/g. R is maximum when theta = 45 degrees.

T = 2u sin(theta)/g ; H = u^2 sin^2(theta)/2g ; R = u^2 sin(2theta)/g

A Carnot cycle has four reversible steps: (1) isothermal expansion at the source temperature T_h (heat Q_h absorbed); (2) adiabatic expansion (temperature falls from T_h to T_c); (3) isothermal compression at the sink temperature T_c (heat Q_c rejected); (4) adiabatic compression (temperature rises from T_c back to T_h). The efficiency is eta = 1 - Q_c/Q_h = 1 - T_c/T_h, where temperatures are in kelvin. Efficiency depends only on the two temperatures and is always less than 1.

eta = 1 - T_c/T_h

Write tan A = sin A/cos A and cot A = cos A/sin A. First term: cos A/(1 - sin A/cos A) = cos A/((cos A - sin A)/cos A) = cos^2 A/(cos A - sin A). Second term: sin A/(1 - cos A/sin A) = sin A/((sin A - cos A)/sin A) = sin^2 A/(sin A - cos A) = -sin^2 A/(cos A - sin A). Adding: [cos^2 A - sin^2 A]/(cos A - sin A) = [(cos A - sin A)(cos A + sin A)]/(cos A - sin A) = cos A + sin A. Hence proved.

cos^2 A - sin^2 A = (cosA - sinA)(cosA + sinA)