tan [i log ((a-ib/a+ib))] is equal to

Question

tan [i log ((a-ib/a+ib))] is equal to (A) ab (B) (2ab/a2-b2) (C) (a2-b2/ab) (D) (2ab/a2+b2)

Tardigrade Admin · Accepted Answer

We have, tan [i log((a-ib/a+ib))] Let a=r cos θ, b=r sin θ ∴ (b/a)=tan θ ldots(i) ∴ tan[i log(e-iθ × e-iθ)] =tan [i log e-2iθ]=tan 2θ (∵ log e=1) =(2 tan θ/1-tan2 θ)=(2(a/b)/1-(b2)a2) =(2ab/a2-b2) (using (i))

Q. $t an [i l o g (\frac{a - ib}{a + ib})]$ is equal to

$ab$

$\frac{2 ab}{a ^{2} - b ^{2}}$

$\frac{a ^{2} - b ^{2}}{ab}$

$\frac{2 ab}{a ^{2} + b ^{2}}$

Q. tan[ilog(a+iba−ib​)] is equal to

ab

a2−b22ab​

aba2−b2​

a2+b22ab​

We have, tan[ilog(a+iba−ib​)] Let a=rcosθ, b=rsinθ ∴ab​=tanθ…(i) ∴tan[ilog(e−iθ×e−iθ)] =tan[iloge−2iθ]=tan2θ(∵loge=1) =1−tan2θ2tanθ​=1−a2b2​2ba​​ =a2−b22ab​ (using (i))

Q. $t an [i l o g (\frac{a - ib}{a + ib})]$ is equal to

$ab$

$\frac{2 ab}{a ^{2} - b ^{2}}$

$\frac{a ^{2} - b ^{2}}{ab}$

$\frac{2 ab}{a ^{2} + b ^{2}}$