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we know cos A = Sin (90 - A) and Sin (A+B) - Sin (A-B) = 2 Sin B Cos A

= Sin[90 - (40-x)] - Sin (50-x) + (cos² 40 + Sin² (90 - 50)) /(Sin² 40 + Cos²(90-50) = Sin (50 +x) - Sin(50 -x) + (cos²40 + Sin²40) /(sin²40 +cos²40) = 2 Sin x Cos 50 + 1 / 1 = 1 + 2 Sin x Cos 50 =============================================== If the question is: Cos (40 - x) - Sin (50 -x) + cos² 40 + (Cos² 50 / Sin² 40) + Sin² 50 = [ Sin (90 - 40 + x) + sin (x - 50) ] + cos² 40 + (Cos²50 / Cos²(90-40)) + Cos² (90-50) = 2 Sin x Cos 50 + 2 Cos² 40 + 1/1 = 2 sin x cos 50 + (1 + cos 80) + 1 = 2 Sin x Cos 50 + 2 + Cos (50 + 30) = 2 Sin x Cos 50 + 2 + √3/2 * cos 50 + 1/2 * Sin 50