If sec θ + tan θ = p, obtain the values of sec θ, tan θ and sin θ

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Q) IF sec θ=(x + 1/4x) then prove that sec θ + tan θ = 2x or 1/2x

Ans:

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

$tan^{2}\theta=sec^{2}\theta-1$

$tan^{2}\theta=(x+\frac{1}{4x})^{2}-1$

$=x^{2}+\frac{1}{16x^{2}}+\frac{1}{2}-1$

$=x^{2}+\frac{1}{16x^{2}}-\frac{1}{2}$

$=(x-\frac{1}{4x})^{2}$

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

$\implies tan\theta=(x-\frac{1}{4x})$

$or$

$tan\theta=-(x-\frac{1}{4x})$

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

When

$tan\theta=(x-\frac{1}{4x})$

$\therefore \sec \theta + \tan \theta = x + \cancel{\frac{1}{4x}} + x - \cancel{\frac{1}{4x}} = 2x$

When

$tan~\theta=-(x-\frac{1}{4x})$

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

$= \cancel{x} + \frac{1}{4x} - \cancel{x} + \frac{1}{4x}$

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

Therefore, value of $\sec \theta + \tan \theta = 2 x~or~\frac{1}{2x}$

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