Question

If P = cos theta + i sin theta, q = cos phi+ i sin phi, show that (p+q)/(p-q)=i tan (theta-phi)/2

Answer 1

Answer:

cos

$\frac{5}{5}$

is your answer

Answer 2

The result is proofed.

As per question we need to find  (p+q)/(p-q)=i tan (theta-phi)/2.

The given data is  P = cos theta + i sin theta, q = cos phi+ i sin phi.

Now let

P = cosθ + i sinθ..............(1)

q=  cos∅+ i sin∅...........(2)

Now we add equation (1) and equation (2) we get

P +q= cosθ + i sinθ +  cos∅+ i sin∅,

       = cosθ +  cos∅ +  i sinθ +  i sin∅,

       =  ( cosθ +  cos∅ ) + i ( sinθ + sin∅ ),

       = 2cos(∅+θ)/2*cos(θ-∅)/2 + i {2sin(∅+θ)/2 *sin(θ-∅)/2}

        = 2{cos(∅+θ)/2*cos(θ-∅)/2 + i {sin(∅+θ)/2 *sin(θ-∅)/2}}.........(3)

Now  subtract the equation (2) from equation (1)

P -q = cosθ + i sinθ -(cos∅+ i sin∅),

       = cosθ - cos∅ + i sinθ- i sin∅

      = ( cosθ - cos∅) +i ( sinθ -sin∅)

     = -2sin(∅+θ)/2 *sin(θ-∅)/2 + i{ 2cos(∅+θ)/2*sin(θ-∅)/2}

     =-2{ sin(∅+θ)/2 *sin(θ-∅)/2 +  i{ cos(∅+θ)/2*sin(θ-∅)/2}}......(4)

Now we divide equation (3) by equation (4)

$\frac{p+q}{p-q}$   $=\frac{ 2{ cos(∅+θ)/2*cos(θ-∅)/2 + i {sin(∅+θ)/2 *sin(θ-∅)/2}}}{ -2{sin(∅+θ)/2 *sin(θ-∅)/2 +i{ cos(∅+θ)/2*sin(θ-∅)/2}}}$

        =i tan(θ-∅)/2

Hence the result is proofed.

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 https://brainly.in/question/23142213?

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