Question:

Multiplicative inverse of the complex number $(\sin\theta, \cos\theta)$ is

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To find the inverse of a complex number $a + ib$, use $\dfrac{a - ib}{a^2 + b^2}$.
Updated On: May 19, 2025
  • $(+\sin\theta, +\cos\theta)$
  • $(\sin\theta, -\cos\theta)$
  • $(\cos\theta, \sin\theta)$
  • $(-\cos\theta, \sin\theta)$
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The Correct Option is B

Solution and Explanation

The complex number is $z = \sin\theta + i\cos\theta$
Its inverse is $\dfrac{1}{z} = \dfrac{\bar{z}}{|z|^2} = \dfrac{\sin\theta - i\cos\theta}{1} = \sin\theta - i\cos\theta$
Hence, in ordered pair form: $(\sin\theta, -\cos\theta)$
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