If cos⁡xa=cos⁡(x+θ)b=cos⁡(x+2θ)c=cos⁡(x+3θ)d then a+cb+d is equal to

# If $\frac{\mathrm{cos}\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{cos}\left(\mathrm{x}+\mathrm{\theta }\right)}{\mathrm{b}}=\frac{\mathrm{cos}\left(\mathrm{x}+2\mathrm{\theta }\right)}{\mathrm{c}}=\frac{\mathrm{cos}\left(\mathrm{x}+3\mathrm{\theta }\right)}{\mathrm{d}}$ then $\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}+\mathrm{d}}$ is equal to

1. A

$\frac{\mathrm{a}}{\mathrm{d}}$

2. B

$\frac{\mathrm{c}}{\mathrm{d}}$

3. C

$\frac{\mathrm{b}}{\mathrm{c}}$

4. D

$\frac{\mathrm{d}}{\mathrm{a}}$

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### Solution:

$\begin{array}{l}\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}+\mathrm{d}}=\frac{\mathrm{cos}\mathrm{x}+\mathrm{cos}\left(\mathrm{x}+2\mathrm{\theta }\right)}{\mathrm{cos}\left(\mathrm{x}+\mathrm{\theta }\right)+\mathrm{cos}\left(\mathrm{x}+3\mathrm{\theta }\right)}\\ =\frac{2\mathrm{cos}\left(\mathrm{x}+\mathrm{\theta }\right)\mathrm{cos}\mathrm{\theta }}{2\mathrm{cos}\left(\mathrm{x}+2\mathrm{\theta }\right)\mathrm{cos}\mathrm{\theta }}=\frac{\mathrm{b}}{\mathrm{c}}\end{array}$