Solution:
We have:
$\sec\text{x} +\tan\text{x} = \text{k}\cdots(1)$
$\Rightarrow\frac{1}{\sec\text{x} + \tan\text{x}}=\frac{1}{\text{k}}$
$\Rightarrow\frac{\sec^2\text{x}-\tan^2\text{x}}{\sec\text{x}+\tan\text{x}} = \frac{1}{\text{k}}$
$\Rightarrow\frac{(\sec\text{x} + \tan\text{x})(\sec\text{x}-\tan\text{x})}{(\sec\text{x} + \tan\text{x})} = \frac{1}{\text{k}}$
$\therefore\sec\text{x} - \tan\text{x} = \frac{1}{\text{k}}\cdots(2)$
Adding (1) and (2):
$2\sec\text{x}= \text{k} + \frac{1}{\text{k}}$
$\Rightarrow 2\sec\text{x} = \frac{\text{k}^2 + 1}{\text{k}}$
$\Rightarrow \sec\text{x} = \frac{\text{k}^2+1}{2\text{k}}$
$\Rightarrow\frac{1}{\cos\text{x}}= \frac{\text{k}^2 + 1}{2\text{k}}$
$\Rightarrow\cos \text{x} = \frac{2\text{k}}{\text{k}^2 + 1}$
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The negation of the statement.
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$\text{T}_{5}$
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$\text{T}_{8}$