Question

Hi there!

Prove that ↑

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Answer 1

Given :-

$\mathsf{\dfrac{Cos\theta + \iota Sin\theta }{Cos\phi + \iota Sin\phi}}$

To show :-

$\mathsf{Cos(\theta - \phi) +\iota Sin(\theta - \phi)}$

Solution:-

• Considering L. H. S

→$\mathsf{ \dfrac{Cos\theta + \iota Sin\theta }{Cos\phi + \iota Sin\phi }\times \dfrac{Cos\phi -\iota Sin \phi}{Cos\phi - \iota Sin\phi}}$

→$\mathsf{ \dfrac{(Cos\theta + \iota Sin\theta )(Cos\phi - \iota Sin \phi )}{Cos^2 \phi -\iota^2 Sin^2\phi}}$

→$\mathsf{\dfrac{Cos\theta .Cos\phi - Cos\theta.\iota Sin\phi+\iota Sin\theta .Cos\phi -\iota^2 Sin\theta .Sin\phi}{Cos^2\phi + Sin^2\phi}}$

→$\mathsf{\dfrac{Cos\theta .Cos\phi + Sin\theta .Sin\phi +\iota (Sin\theta .Cos\phi -Sin\phi.Cos\theta)}{1}}$

→$\mathsf{ Cos(\theta -\phi) + \iota Sin(\theta - \phi)}$

hence, proved...

Formula used :-

$\mathsf{{Cos(A-B)= CosA. CosB +SinA. SinB}}$

$\mathsf{{Sin(A-B) = SinA. CosB -SinB.CosA}}$

Answer 2

Heya!

Refer to the attachment!