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# If $\sec \theta = x + \dfrac{1}{{4x}},$ then prove that, $\sec \theta + \tan \theta = 2x$ or $\dfrac{1}{{2x}}.$

Last updated date: 14th Jul 2024
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Hint: Here we will use the identity for the secant and tangent relation by using the relation $1 + {\tan ^2}x = {\sec ^2}x$ and will find the tangent angle and then find the value of sum of secant and tangent and will prove the value given in the right hand side of the equation.

Given that: $\sec \theta = x + \dfrac{1}{{4x}}$
Simplify the above relation finding the LCM (least common multiple)
$\sec \theta = \dfrac{{4{x^2} + 1}}{{4x}}$ ….. (A)
Use the identity, $1 + {\tan ^2}x = {\sec ^2}x$
Make the tangent the subject, when you move any term from one side to another the sign of the term also changes. Positive term changes to the negative and vice-versa.
${\tan ^2}x = {\sec ^2}x - 1$
Place the value of A in the above equation -
${\tan ^2}x = {\left( {\dfrac{{4{x^2} + 1}}{{4x}}} \right)^2} - 1$
Simplify the above expression finding the square of the terms –
${\tan ^2}x = \left( {\dfrac{{16{x^4} + 8{x^2} + 1}}{{16{x^2}}}} \right) - 1$
Take LCM (least common multiple) for the terms in the above expression –
${\tan ^2}x = \left( {\dfrac{{16{x^4} + 8{x^2} + 1 - 16{x^2}}}{{16{x^2}}}} \right)$
Combine the like terms together –
${\tan ^2}x = \left( {\dfrac{{16{x^4} + \underline {8{x^2} - 16{x^2}} + 1}}{{16{x^2}}}} \right)$
When you subtract a bigger term from the positive smaller number then the result is negative.
${\tan ^2}x = \left( {\dfrac{{16{x^4} - 8{x^2} + 1}}{{16{x^2}}}} \right)$
The above term can be expressed as the whole square of the terms.
${\tan ^2}x = {\left( {\dfrac{{4{x^2} - 1}}{{4x}}} \right)^2}$
Take square root on both the sides of the equation –
$\sqrt {{{\tan }^2}x} = \sqrt {{{\left( {\dfrac{{4{x^2} - 1}}{{4x}}} \right)}^2}}$
Square and square root cancel each other
$\tan x = \pm \left( {\dfrac{{4{x^2} - 1}}{{4x}}} \right)$
We have two case –
When, $\tan x = \left( {\dfrac{{4{x^2} - 1}}{{4x}}} \right)$
Now, place values in $\sec \theta + \tan \theta$
$\sec \theta + \tan \theta = \dfrac{{4{x^2} + 1}}{{4x}} + \dfrac{{4{x^2} - 1}}{{4x}}$
When denominators are the same, combine the numerator.
$\sec \theta + \tan \theta = \dfrac{{4{x^2} + 1 + 4{x^2} - 1}}{{4x}}$
Like terms with the same value and opposite sign cancels each other.
$\sec \theta + \tan \theta = \dfrac{{8{x^2}}}{{4x}}$
Common factors from the numerator and the denominator cancel each other.
$\sec \theta + \tan \theta = 2x$ …… (A)
Second case: When, $\tan x = - \left( {\dfrac{{4{x^2} - 1}}{{4x}}} \right)$
Now, place values in $\sec \theta + \tan \theta$
$\sec \theta + \tan \theta = \dfrac{{4{x^2} + 1}}{{4x}} - \dfrac{{4{x^2} - 1}}{{4x}}$
When denominators are the same, combine the numerator.
$\sec \theta + \tan \theta = \dfrac{{4{x^2} + 1 - 4{x^2} + 1}}{{4x}}$
Like terms with the same value and opposite sign cancels each other.
$\sec \theta + \tan \theta = \dfrac{2}{{4x}}$
Common factors from the numerator and the denominator cancel each other.
$\sec \theta + \tan \theta = \dfrac{1}{{2x}}$ …… (B)
(A) And (B) are the required solutions.

Note: Always remember that the square of positive or the negative term always gives positive term. When there is a negative sign outside the term then the sign of the terms inside changes. Positive terms become negative and negative terms become positive terms. Be careful about the sign convention.