Question

if A+iB ctan(x+iy) prove that tan2x=2CA/C2-A2cB2​

Answer 1

Step-by-step explanation:

tan(x+y) = (tanx+tany)/1-tanx.tany

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Answer 2

Answer:

If $A+i B=c \tan (x+i y)$ then $tan2x=\frac{2CA}{c^{2}-A^{2} -B^{2} }$.

Step-by-step explanation:

Given :

$A+i B=c \tan (x+i y)$

To find :  

$tan2x=\frac{2CA}{c^{2}-A^{2} -B^{2} }$

Step 1

$\tan (x+i y)=\frac{A+i B}{C}$................(1)

$\tan (x-i y)=\frac{A-i B}{C}$................(2)

Take $tan 2x$ as,

$\tan 2 x=\tan [(x+i y)+(x-i y)]$

$=\left[\frac{\tan (x+i y)+\tan (x-i y)}{1-\tan (x+i y) \cdot \tan (x-i y)}\right]$

Step 2

Then we get,

$\tan 2 x=\frac{A+i B}{C}+\frac{A-i B}{C}$

$= 1-\left(\frac{A-i B}{C}\right) \cdot\left(\frac{A+i B}{C}\right)$

$=\frac{(A+iB+A-iB)C^{2} }{[C^{2}-(A-i B)(A+i B)]\right }$

Step 3

Cancelling the same terms of numerator and denominator, we get

$tan2x=\frac{2CA}{c^{2}-A^{2} -B^{2} }$.

Therefore, we get

$tan2x=\frac{2CA}{c^{2}-A^{2} -B^{2} }$.

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