If the value of e -2 q i tan -1 p i - p - i - p q - R- where q - p - I - i -1- Then the value of R 3 is -A- e -3 iB- e3 iC- e 3D- 1

R=e^ 2q i(tan^ 1p)(i pi+p)^q Assuming tan^ 1p=θ ⇒tanθ =p So we can write, ⇒ R= e^ 2q iθ( i tanθi+tanθ)^q ⇒ R= e^ 2q iθ( i+i^2tanθi i^2tanθ)^q ⇒ R= e^ 2q iθ(1+ ...

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Byju's Answer

Standard XII

Mathematics

Sin(A+B)Sin(A-B)

If the value ...

Question

If the value of $e^{- 2 q i ({tan}^{- 1} p)} {(\frac{i - p}{i + p})}^{q} = R$ , where $q, p \in I & i = \sqrt{- 1}$ . Then the value of $R^{3}$ is :

e^{- 3 i}

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e^{3 i}

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e^{3}

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1

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Solution

The correct option is D $1$
$R = e^{- 2 q i ({tan}^{- 1} p)} {(\frac{i - p}{i + p})}^{q}$
Assuming ${tan}^{- 1} p = θ$
$\Rightarrow tan θ = p$
So we can write,
$\Rightarrow R = e^{- 2 q i θ} {(\frac{i - tan θ}{i + tan θ})}^{q} \Rightarrow R = e^{- 2 q i θ} {(\frac{i + i^{2} tan θ}{i - i^{2} tan θ})}^{q}$
$\Rightarrow R = e^{- 2 q i θ} {(\frac{1 + i tan θ}{1 - i tan θ})}^{q}$
$\Rightarrow R = e^{- 2 q i θ} {(\frac{e^{i θ}}{e^{- i θ}})}^{q} \Rightarrow R^{3} = 1$

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If the value of e−2q i(tan−1p)(i−pi+p)q=R, where q,p∈I & i=√−1. Then the value of R3 is :

The correct option is D 1R=e−2q i(tan−1p)(i−pi+p)q Assuming tan−1p=θ ⇒tanθ=p So we can write, ⇒R=e−2q iθ(i−tanθi+tanθ)q⇒R=e−2q iθ(i+i2tanθi−i2tanθ)q ⇒R=e−2q iθ(1+itanθ1−itanθ)q ⇒R=e−2q iθ(eiθe−iθ)q⇒R3=1

If the value of $e^{- 2 q i ({tan}^{- 1} p)} {(\frac{i - p}{i + p})}^{q} = R$ , where $q, p \in I & i = \sqrt{- 1}$ . Then the value of $R^{3}$ is :