Question

If $secA = x + \dfrac{1}{{4x}},x \ne 0$ then find the value of $(\operatorname{Sec} A + \operatorname{Tan} A)$.

Answer 1

First of all find the’$secA$’from the given condition, then just reverse it to get the value of$\operatorname{Cos} A$.
By using identity: $si{n^2}A + co{s^2}A = 1$, find $\operatorname{Sin} A$.
Now use the relationship between$sinA,cosA$and $tanA$to find$secA$.$$$$
Then substitute the values of both $tanA$and$secA$in the given equation.
$\therefore (tanA + secA) = 2x$

Complete step-by-step answer:
We have $secA = x + \dfrac{1}{{4x}},x \ne 0$
By taking L.C.M as $4x$, we get
$secA = x + \dfrac{1}{{4x}} = \dfrac{{4{x^2} + 1}}{{4x}}$
Therefore, $cosA = \dfrac{1}{{secA}} = \dfrac{{4x}}{{4{x^2} + 1}}$.
For any acute angle$A$, we have identity: $si{n^2}A + co{s^2}A = 1$.
Hence, $sinA = \sqrt {1 - co{s^2}A}$
Simply substitute the value of $\operatorname{Cos} A$in the above equation, we get
$sinA = \sqrt {1 - {{\left( {\dfrac{{4x}}{{4{x^2} + 1}}} \right)}^2}}$
$sinA = \sqrt {1 - \dfrac{{16{x^2}}}{{{{(4{x^2} + 1)}^2}}}}$
Taking L.C.M as ${(4{x^2} + 1)^2}$in the denominator
$sinA = \sqrt {\dfrac{{{{(4{x^2} + 1)}^2} - 16{x^2}}}{{{{(4{x^2} + 1)}^2}}}}$
$\Rightarrow sinA = \sqrt {\dfrac{{{{(4{x^2} - 1)}^2}}}{{{{(4{x^2} + 1)}^2}}}}$
We can write the above equation after removing the square as
$sinA = \pm \dfrac{{(4{x^2} - 1)}}{{(4{x^2} + 1)}}$
If $sinA = \dfrac{{(4{x^2} - 1)}}{{(4{x^2} + 1)}}$then
We have
$secA + tanA = \dfrac{{4{x^2} + 1}}{{4x}} + \dfrac{{4{x^2} - 1}}{{4x}}$
After solving, we get
$secA + tanA = 2x$
If $sinA = - \dfrac{{(4{x^2} - 1)}}{{(4{x^2} + 1)}}$then
$secA + tanA = \dfrac{{4{x^2} + 1}}{{4x}} - \dfrac{\begin{gathered} \\\ (4{x^2} - 1) \\\ \end{gathered} }{{4x}}$
On further solving we get,

$secA + tanA = \dfrac{1}{{2x}}$

Note: The trigonometric ratios of an acute angle in a right triangle express the relationship between the angles and the length of its sides.
Tangent of $\angle A = \dfrac{{Perpendicular}}{{Base}}$ i.e. $tanA = \dfrac{P}{B}$ ,
Cotangent of $\angle A = \dfrac{1}{{tanA}} = \dfrac{B}{P}$ $\angle A = \dfrac{1}{{tanA}} = \dfrac{B}{P}$,
Secant of$\angle A = \dfrac{1}{{cosA}} = \dfrac{{Hypotenuse}}{{Base}} = \dfrac{H}{B}$and
For any acute angle $A$, we have identity: $si{n^2}A + co{s^2}A = 1$