Determine the smallest positive value of $x$ (in degrees) for which $\tan (x + 100 °) = \tan (x + 50 °) \tan (x) \tan (x - 50 °)$ .

$30 °$

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$45 °$

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$60 °$

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$75 °$

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Solution

The correct option is A
$30 °$

Finding the smallest positive value of $x$
$\tan (x + 100 °) = \tan (x + 50 °) \tan (x) \tan (x - 50 °)$
$\frac{\tan (x + 100 °)}{\tan (x - 50 °)} = \tan (x + 50 °) \tan (x)$
$\frac{\sin (x + 100 °) \cos (x - 50 °)}{\cos (x + 100 °) \sin (x - 50 °)} = \frac{\sin (x + 50 °) \sin (x)}{\cos (x + 50 °) \cos (x)}$
Applying componendo and dividendo we get,
$\frac{\sin (2 x + 50 °)}{\sin (150 °)} = \frac{- \cos (50 °)}{\cos (2 x + 50 °)}$
$\sin (2 x + 50 °) \times \cos (2 x + 50 °) = - \cos (50 °) \times \sin (150 °)$
$\sin (4 x + 100 °) = - \sin (40 °)$
$\sin (4 x + 100 °) = \sin (- 40 °)$
$4 x + 100 ° = - 40 °$ {If $\sin (y) = \sin (x) \Rightarrow y = x$ }
$4 x + 100 ° = πn - {(- 1)}^{n} 40 °$
Substitute $n = 1$ , smallest positive value of $x$ is given by,
$4 x = 120 °$
$x = 30 °$
Hence, option A is correct.

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