Question

Complex numbers class 11
Please help
I have solved till last step but the answer is not correct
The answer is (b) option ​

Answer 1

Answer:

$\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=2\:cos(\alpha+\beta+\gamma+\delta)$

option (B) is correct

Step-by-step explanation:

$\text{Given:}$

$a=cos2\alpha+i\:sin2\alpha=e^{i2\alpha}$

$b=cos2\beta+i\:sin2\beta=e^{i2\beta}$

$c=cos2\gamma+i\:sin2\gamma=e^{i2\gama}$

$d=cos2\delta+i\:sin2\delta=e^{i2\delta}$

Now,

$abcd=e^{i2\alpha}e^{i2\beta}e^{i2\gamma}e^{i2\delta}$

$\implies\:abcd=e^{i2(\alpha+\beta+\gamma+\delta)}$

$\implies\:abcd=cos\:2(\alpha+\beta+\gamma+\delta)+i\:sin\:\:2(\alpha+\beta+\gamma+\delta)$

$\implies\:\sqrt{abcd}=[cos\:2(\alpha+\beta+\gamma+\delta)+i\:sin\:\:2(\alpha+\beta+\gamma+\delta)]^{1/2}$

using Demovire's theorem

$\boxed{(cos\theta+i\:sin\theta)^n=cos\:n\theta+i\:sin\:n\theta}$

$\sqrt{abcd}=[cos\:2\times\frac{1}{2}(\alpha+\beta+\gamma+\delta)+i\:sin\:2\timesfrac{1}{2}(\alpha+\beta+\gamma+\delta)]$

$\implies\:\sqrt{abcd}=cos(\alpha+\beta+\gamma+\delta)+i\:sin(\alpha+\beta+\gamma+\delta)$..............(1)

$\frac{1}{\sqrt{abcd}}=\frac{1}{e^{i(\alpha+\beta+\gamma+\delta)}}$

$\frac{1}{\sqrt{abcd}}=e^{-i(\alpha+\beta+\gamma+\delta)}$

$\frac{1}{\sqrt{abcd}}=cos(\alpha+\beta+\gamma+\delta)-i\:sin(\alpha+\beta+\gamma+\delta)$............(2)

Adding (1) and (2), we get

$\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=cos(\alpha+\beta+\gamma+\delta)+i\:sin(\alpha+\beta+\gamma+\delta)+cos(\alpha+\beta+\gamma+\delta)-i\:sin(\alpha+\beta+\gamma+\delta)$

$\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=cos(\alpha+\beta+\gamma+\delta)++cos(\alpha+\beta+\gamma+\delta)$

$\therefore\boxed{\bf{\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=2\:cos(\alpha+\beta+\gamma+\delta)}}$