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if alpha and beta are the roots of the equation 3cosx+4sinx=5 then cot(alpha+beta) equals
3cosx + 4sinx = Asin(x + Ф) Multiply and divide with = 5 in LHSe.g., (3cosx + 4sinx)/5 = Asin(x + Ф)⇒5(3/5.cosx + 4/5.sinx) = Asin(x + Ф) Let cosФ= 3/5 then, sinФ = 4/5Now, tanФ = sinФ/cosФ = 4/3⇒ Ф = tan⁻¹(4/3) ----(1) And 5(3/5.cosx + 4/5.sinx)= 5(sinФ.cosx + cosФ.sinx) = Asin(x + Ф)⇒ 5sin(x + Ф) = Asin(x + Ф)From equation (1),⇒ 5sin(x + tan⁻¹(4/3)) = Asin(x + Ф)Compare both sides,A = 5 and Ф = tan⁻¹(4/3)
Dear studentThe simplification isSin ( x + 37 ) = 1= sin ( 90+ 2n*pi)You can solve accordingly by evaluating alpha and betaRegards
note that aruns soln doesnt answer the ques.we use the standard formulas:cosx= (1 – t^2)/(1+t^2) and sinx= 2t/(1+t^2) where t= tan(x/2).so eqn becomes3(1 – t^2)+4(2t)= 5(1+t^2)or 8t^2 – 8t + 2= 0 or 4t^2 – 4t + 1= 0or (2t – 1)^2= 0or t= tan(x/2)= ½ so that tan(p)= tan(q)= ½ where p= alpha/2 and q= beta/2.cot(alpha+beta)= 1/tan(alpha+beta)= 1/tan(2p + 2q)= (1 – tan2p*tan2q)/(tan2p + tan2q)now, tan2p= 2tanp/(1 – tan^2p)= 4/3. similarly tan2q= 4/3hence, cot(alpha+beta)= (1 – tan2p*tan2q)/(tan2p + tan2q)= – 7/24kindly approve :))
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