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Question
If \[\tan x = t\] then \[\tan 2x + \sec 2x =\]
Options
- \[\frac{1 + t}{1 - t}\]
- \[\frac{1 - t}{1 + t}\]
- \[\frac{2t}{1 - t}\]
- \[\frac{2t}{1 + t}\]
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Solution
\[\frac{1 + t}{1 - t}\]
\[\tan 2x + \sec 2x = \frac{2 \tan x}{1 - \tan^2 x} + \frac{1 + \tan^2 x}{1 - \tan^2 x}\]
\[ = \frac{2\tan x + 1 + \tan^2 x}{1 - \tan^2 x}\]
\[ = \frac{\left( 1 + \tan x \right)^2}{1 - \tan^2 x}\]
\[ = \frac{\left( 1 + \tan x \right)\left( 1 + \tan x \right)}{\left( 1 + \tan x \right)\left( 1 - \tan x \right)}\]
\[ = \frac{1 + \tan x}{1 - \tan x}\]
\[ = \frac{1 + t}{1 - t} \left[ \tan x = t \left( \text{ given } \right) \right]\]
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