Question

If sec tetha=x+1/4x find the value of sec tetha+tan tetha

Answer 1

$<u> <b> Here's your answer </b> </u>$

sec = x + $\bf\frac{1}{4x}$ (given)

As we know

$1 + tan^2\thita = sec^2\thita$

=>tan² = (x + $\bf\frac{1}{4x}$)² - 1

=>tan² = x² + $\bf\frac{1}{16x}$ + 2(x)($\bf\frac{1}{4x}$)

=>x² + $\bf\frac{1}{16x}$ - $\bf\frac{1}{2}$

=> (X - $\bf\frac{1}{4x}$)²

thus,$\boxed{\bf {tan = (x- \bf\frac{1}{4x^2}) or -(x -\bf\frac{1}{4x^2})}}$

$\bf\underline{Now\: putting\: the\: values}$

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

= 2x

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

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

$<marquee>Hope it helps you :)</marquee>$

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