If  tan x =1213, then evaluate the value of  2sin⁡xcos⁡xcos2⁡x-sin2⁡x.

# If , then evaluate the value of  $\frac{2\mathit{sin}x\mathit{cos}x}{{\mathit{cos}}^{2}x-{\mathit{sin}}^{2}x}$.

1. A
312/25
2. B
350/40
3. C
150/42
4. D
450/41

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### Solution:

Given,
$\frac{2\mathit{sin}x\mathit{cos}x}{{\mathit{cos}}^{2}x-{\mathit{sin}}^{2}x}$          …(1)
We know that,
$\mathit{sin}2a=2\mathit{sin}a\mathit{cos}a$  and $\mathit{cos}2a={\mathit{cos}}^{2}a-{\mathit{sin}}^{2}a$
Apply these identities in equation 1,
$\frac{2\mathit{sin}x\mathit{cos}x}{{\mathit{cos}}^{2}x-{\mathit{sin}}^{2}x}$
$=\frac{\mathit{sin}2x}{\mathit{cos}2x}$
$=\mathit{tan}2x$
$=\frac{2\mathit{tan}x}{1-{\mathit{tan}}^{2}x}$                         (Since, $\mathit{tan}2a=\frac{2\mathit{tan}a}{1-{\mathit{tan}}^{2}a}$)
$=\frac{2\frac{12}{13}}{1-{\left(\frac{12}{23}\right)}^{2}}$                        (Given $\mathit{tan}x=\frac{12}{13}$)
$=\frac{\frac{24}{13}}{1-\frac{144}{169}}$
$=\frac{\frac{24}{13}}{\frac{169-144}{169}}$
$=\frac{\frac{24}{13}}{\frac{25}{169}}$
$=\frac{24}{13}×\frac{169}{25}$
$=\frac{24}{1}×\frac{13}{25}$
$=\frac{312}{25}$
$=\frac{312}{25}$
Hence, option 1 is correct.

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