If cos⁡θ=12x+1x then 12×2+1×2 is equal to

# If $\mathrm{cos}\mathrm{\theta }=\frac{1}{2}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)$ then $\frac{1}{2}\left({\mathrm{x}}^{2}+\frac{1}{{\mathrm{x}}^{2}}\right)$ is equal to

1. A

sin 2$\mathrm{\theta }$

2. B

cos 2$\mathrm{\theta }$

3. C

tan 2$\mathrm{\theta }$

4. D

None of these

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### Solution:

Given that, $\mathrm{cos}\mathrm{\theta }=\frac{1}{2}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)$

$⇒\mathrm{x}+\frac{1}{\mathrm{x}}=2\mathrm{cos}\mathrm{\theta }----\left(\mathrm{i}\right)$

We know that,

$\begin{array}{l}{\mathrm{x}}^{2}+\frac{1}{{\mathrm{x}}^{2}}={\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^{2}-2\\ =\left(2\mathrm{cos}\mathrm{\theta }{\right)}^{2}-2=4{\mathrm{cos}}^{2}\mathrm{\theta }-2\\ =2\mathrm{cos}2\mathrm{\theta }\\ \therefore \frac{1}{2}\left({\mathrm{x}}^{2}+\frac{1}{{\mathrm{x}}^{2}}\right)=\frac{1}{2}×2\mathrm{cos}2\mathrm{\theta }=\mathrm{cos}2\mathrm{\theta }\end{array}$

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