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if $\sec\theta = x + \frac{1}{4x}$ $x\neq0,$ find $sec\theta+tan\theta$

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Given: $\sec\theta = x + \frac{1}{4x}$

To Find: $\sec\theta +\tan \theta$

We know that $\Rightarrow \tan^2\theta = \sec^2\theta - 1\quad (\text{trignometric identity})$

$\Rightarrow \tan^2\theta =\left(x + \frac{1}{4x} \right )^2 - 1\quad (\text{trignometric identity})$

$\begin{align*} \Rightarrow \tan^2\theta & =x^2 + \frac{1}{(4x)^2} + 2 \times x\times \frac{1}{4x} - 1 \\ & = x^2 + \frac{1}{(4x)^2} + \frac{1}{2} - 1 \\ & = x^2 + \frac{1}{(4x)^2} - \frac{1}{2} \\ & = x^2 + \frac{1}{(4x)^2} - \frac{1}{2}\times \frac{2}{2}\frac{x}{x} \\ &= x^2 + \frac{1}{(4x)^2} - 2\times \frac{1}{4x}\times x \\ & = \left(x - \frac{1}{4x} \right )^2 \qquad(\because (a-b)^2 = a^2 + b^2-2ab)\end{align*}$

$\Rightarrow \tan\theta = \sqrt{\left(x - \frac{1}{4x} \right )^2}$

$\Rightarrow \tan\theta = \pm \left(x - \frac{1}{4x} \right )$

$\Rightarrow \tan\theta = x - \frac{1}{4x}$ OR $\Rightarrow \tan\theta = -x + \frac{1}{4x}$

When $\Rightarrow \tan\theta = x- \frac{1}{4x}$

$\sec\theta +\tan \theta = x + \frac{1}{4x} + x -\frac{1}{4x}$

$= 2x$

When $\Rightarrow \tan\theta = -x + \frac{1}{4x}$

$\sec\theta +\tan \theta = x + \frac{1}{4x} - x +\frac{1}{4x}$

$= \frac{1}{2x}$

$\Rightarrow \sec\theta +\tan \theta = 2x$ OR $\frac{1}{2x}$

Posted by

Safeer PP

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