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# if 2CosA = x+1//x and 2CosB = y+1/y then find 2cos(A+B)(A) x/y +y/x(B) x/y-y/x(C) 2y/x(D) 2x/y

divyanshu
27 Points
6 years ago
sorry there is a mistake in my typing in question ie 2cos A= x + 1/x

divyanshu
27 Points
6 years ago

Y RAJYALAKSHMI
45 Points
6 years ago
Solve these as quadratic equations in x & y so that you get x = cosA + i sinA & cosA – i sinA;
y = cosB + i sinB & cosB – isinB
If x = cosA + i sinA then 1/x = cosA – i sinA ;
y = cosB + i sinB then 1/y  cosB – isinB
x + 1/x = 2cosA => cosA = 1/2(x + 1/x)
x – 1/x = 2i sinA => sinA = 1/2i(x – 1/x)
y + 1/y = 2cosB => cosB = 1/2(y + 1/y)
y – 1/y = 2i sinB => sinB = 1/2i(y – 1/y)
2cos(A + B) = 2(cosA cosB – sinA sinB)
= 2[1/2(x + 1/x) * 1/2(y + 1/y) – 1/2i(x – 1/x) * 1/2i(y – 1/y)]
= 2[1/4(xy + x/y + y/x + 1/xy + xy -x/y – y/x + 1/xy)]
= xy + 1/xy

Ans: xy + 1/xy

Kaustubh Nayyar
27 Points
6 years ago
hey the ques you have given is not correct the answer for the ques you have given is xy + 1/xy
The answer for 2cos ( A – B ) is x/y + y/x   This can proved by the similar method of using x and y in terms of cosA , sinA, cos B ,sinB