If tan-x-1-4 x- then sec theta-tan itheta is equal to

The correct option is A 2x, 1/2x (a) 2x, 1/2x We have, tan θ =x 1/4x ⇒ sec^2 θ =1+tan^2 θ ⇒ sec^2 θ =1+ ( x 1/4x )^2 ⇒ sec^2 θ =x^2 +1/16x^2+1/2 ⇒ sec^2 θ = ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

If tanθ=x-1/4...

Question

If $t a n θ = x - \frac{1}{4 x}$ , then sec \theta-tan \theta is equal to

$- 2 x, \frac{1}{2 x}$

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$- \frac{1}{2 x}, 2 x$

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

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Solution

The correct option is A
$- 2 x, \frac{1}{2 x}$

$(a) - 2 x, \frac{1}{2 x}$

We have,

$t a n θ = x - \frac{1}{4 x}$

$\Rightarrow s e c^{2} θ = 1 + t a n^{2} θ$

$\Rightarrow s e c^{2} θ = 1 + {(x - \frac{1}{4 x})}^{2}$

$\Rightarrow s e c^{2} θ = x^{2} + \frac{1}{16 x^{2}} + \frac{1}{2}$

$\Rightarrow s e c^{2} θ = {(x + \frac{1}{4 x})}^{2}$

$∴ s e c θ = \pm (x + \frac{1}{4 x})$

$\Rightarrow s e c θ - t a n θ$

$= (x + \frac{1}{4 x}) - (x - \frac{1}{4 x}) o r - (x + \frac{1}{4 x}) - (x - \frac{1}{4 x})$

$= x + \frac{1}{4 x} x + \frac{1}{4 x} o r x - \frac{1}{4 x} x + \frac{1}{4 x}$

$= \frac{2}{4 x} o r - 2 x$

$= \frac{1}{2 x} o r - 2 x$

Hence $(a) - 2 x, \frac{1}{2 x}$

Suggest Corrections

If tanθ=x−14x, then sec \theta-tan \theta is equal to

If $t a n θ = x - \frac{1}{4 x}$ , then sec \theta-tan \theta is equal to