IF Sec θ=(x+1/4x) then prove that secθ +tanθ =2x or 1/2x

Question

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

Sapling Academy · Accepted Answer

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

$tan^{2}	heta=sec^{2}	heta-1$

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

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

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

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

$	an 	heta = \pm \left(x - \frac{1}{4x}ight)$

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

$or$

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

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

When

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

$	herefore \sec 	heta + 	an 	heta = x + \cancel{\frac{1}{4x}} + x - \cancel{\frac{1}{4x}} = 2x$

When

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

$	herefore \sec 	heta + 	an 	heta = x + \frac{1}{4x} - (x - \frac{1}{4x})$

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

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

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

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