# Prove that Log [ Sin ( X + I Y ) Sin ( X − I Y ) ] = 2 Tan − 1 ( Cot X Tanh Y ) - Applied Mathematics 1

Prove that log [sin(x+iy)/sin(x-iy)]=2tan^-1 (cot x tanhy)

#### Solution

Consider, log[sin(x+iy)]=log[sin x cos(iy) + cos x sin (iy)]
∴log[sin(x+iy)]=log[sin x cosh y + icos x sinh y] {∵cos(ix)=cosh x; sin (ix)=isinh x;}

∴ Log [sin(x+iy)]1/2log[sin^2 xcosh^2 y+cos^2 x sinh^2y]+itan^-1|(cosxsinhy)/(sin xcosh y)|

{∵ log (x-iy)=1/2 log (x^2+y^2)+itan^-1|y/x|}

∴ log[sin(x+iy)]1/2log [sin^2x cosh^2 y+cos^2xx sinh^2 y]+|cot x tanh y|

Taking Conjugate, log[sin(x-iy)=1/2log [sin^2 x cosh^2 y +cos^2 x sinh^2 y]-itan^1|cotx tanh y|

Now, log [sin(x+iy)/sin(x-iy)]= log [sin(sin(x-iy)]

= {1/2 log [sin2 x cosh2y + cos2 x sinh2 y]+itan-1|cot x tanh y|}- {1/2 log [sin2 x cosh 2 y+ cos 2 x sinh2 y]-itan-1|cot x tanhy|}

= 2 itan^-1(cot x tanh y)

∴ log [sin(x+iy)/sin(x-iy)]=2itan^-1 (cot x tanh y)

Concept: Expansion of 𝑒^𝑥 , sin(x), cos(x), tan(x), sinh(x), cosh(x), tanh(x), log(1+x), 𝑠𝑖𝑛−1 (𝑥),𝑐𝑜𝑠−1 (𝑥),𝑡𝑎𝑛−1 (𝑥)
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