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Q.

$\tan 100 ° + \tan 125 ° + \tan 100 ° \tan 125 ° =$

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a

\frac{1}{\sqrt{3}}

b

- 1

c

\frac{1}{2}

d

1

answer is A.

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Detailed Solution

\tan 100 ° + \tan 125 ° + \tan 100 ° \tan 125 °

$\Rightarrow \tan (100 + 125) = \tan 225$

$\Rightarrow \tan (180 + 45) = \frac{\tan 100 + \tan 125}{1 - \tan 100 \tan 125}$

$\Rightarrow 1 = \frac{\tan 100 + \tan 125}{1 - \tan 100. \tan 125}$

$\Rightarrow 1 - \tan 100 \tan 125 = \tan 100 + \tan 125$

$\Rightarrow$ $1 = \tan 100 + \tan 125 + \tan 100 \tan 125$

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