2cosQ+2√2=3secQ find Q solve step by step and give ans
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( r MEANS ROOTS….. SO DON\’T GET CONFUSED )2cosQ +2r2 = 3secQ2cosQ + 2r2 = 3/ cosQ2cos^2Q + 2r2cosQ = 3(Multiplying the eq. by r2)2r2cos^2Q + 4cosQ -3r2 = 02r2cos^2Q – 2cosQ + 6cosQ – 3r2 = 0( by factorization we get )cosQ = 1/r2 and cosQ = -3r2/2 (u can find it out and get same result)Since the value of cosQ can\’t be negative Therefore the actual value of cosQ is 1/r2And hence Q is 45°
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