Question

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Answer 1

Answer

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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

Answer 2

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your answer is in the attachment⚠️❣️⚠️