given tan A = 3/4, show that cos2A - sin2A/cosA = 1 - tan2A/secA - Maths - Introduction to Trigonometry

Question

given tan A = 3/4, show that cos2A - sin2A/cosA = 1 - tan2A/secA

Manbar Singh · Accepted Answer

We have,tan A = 34 = PerpendicularPbaseBnow, hypotenuseH = perpendicular2+base2 = 32+42 = 5Now, cos A = BH = 45sin A = PH = 35sec A = 1cos A = 54now, LHS = cos2A - sin2Acos A = 16/25 - 9/254/5 = 725 × 54 = 720now, RHS = 1-tan2Asec A = 1 - 91654 = 716 × 45 = 720Hence, LHS = RHS