Question

If the trigonometric function can be expressed as, ${\text{sec }}\theta {\text{ = x + }}\dfrac{1}{{4{\text{x}}}}$ then prove that the equation, ${\text{sec }}\theta {\text{ + tan }}\theta {\text{ = 2x or }}\dfrac{1}{{{\text{2x}}}}$.

Answer 1

Hint: In order to prove that, ${\text{sec }}\theta {\text{ + tan }}\theta {\text{ = 2x or }}\dfrac{1}{{{\text{2x}}}}$, we need to express the given trigonometric function, ${\text{sec }}\theta {\text{ = x + }}\dfrac{1}{{4{\text{x}}}}$ in terms of tan function using the appropriate identity to find the value of tan θ. Also the algebraic term of the form can be expressed as, ${\left( {{\text{a - b}}} \right)^2} = {{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}$ and ${\left( {{\text{a + b}}} \right)^2} = {{\text{a}}^2} + {{\text{b}}^2} + 2{\text{ab}}$

Complete step by step solution:
Given data,
${\text{sec }}\theta {\text{ = x + }}\dfrac{1}{{4{\text{x}}}}$

To prove,
${\text{sec }}\theta {\text{ + tan }}\theta {\text{ = 2x or }}\dfrac{1}{{{\text{2x}}}}$

We know the trigonometric identity which states,
${\text{ta}}{{\text{n}}^2}\theta = {\text{ se}}{{\text{c}}^2}\theta {\text{ - 1}}$.

Using the value of the given sec θ, let us compute the value of tan θ, as follows:
$
   \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{ }}{\left( {{\text{x + }}\dfrac{1}{{4{\text{x}}}}} \right)^2}{\text{ - 1}} \\
   \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{ }}{{\text{x}}^2}{\text{ + }}\dfrac{1}{{16{{\text{x}}^2}}}{\text{ + 2}}{\text{.x}}{\text{.}}\dfrac{1}{{4{\text{x}}}}{\text{ - 1}} \\
   \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{ }}{{\text{x}}^2}{\text{ + }}\dfrac{1}{{16{{\text{x}}^2}}}{\text{ + }}\dfrac{1}{2}{\text{ - 1}} \\
$

$
   \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{ }}{{\text{x}}^2}{\text{ + }}\dfrac{1}{{16{{\text{x}}^2}}}{\text{ - }}\dfrac{1}{2} \\
   \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{ }}{\left( {{\text{x - }}\dfrac{1}{{4{\text{x}}}}} \right)^2} \\
   \Rightarrow {\text{tan}}\theta = {\text{ }}\left( {{\text{x - }}\dfrac{1}{{4{\text{x}}}}} \right){\text{ or - }}\left( {{\text{x - }}\dfrac{1}{{4{\text{x}}}}} \right) \\
$

Now we have the values of sec θ and tan θ, let us compute the value of sec θ + tan θ for both the values of tan θ:
${\text{sec }}\theta {\text{ = x + }}\dfrac{1}{{4{\text{x}}}}$
${\text{tan}}\theta = {\text{ }}\left( {{\text{x - }}\dfrac{1}{{4{\text{x}}}}} \right){\text{ or - }}\left( {{\text{x - }}\dfrac{1}{{4{\text{x}}}}} \right)$
$
   \Rightarrow {\text{sec}}\theta {\text{ + tan}}\theta = {\text{ x + }}\dfrac{1}{{4{\text{x}}}} + \left( {{\text{x - }}\dfrac{1}{{4{\text{x}}}}} \right){\text{ or x + }}\dfrac{1}{{4{\text{x}}}}{\text{ - }}\left( {{\text{x - }}\dfrac{1}{{4{\text{x}}}}} \right) \\
   \Rightarrow {\text{sec}}\theta {\text{ + tan}}\theta = {\text{ x + }}\dfrac{1}{{4{\text{x}}}} + {\text{x - }}\dfrac{1}{{4{\text{x}}}}{\text{ or x + }}\dfrac{1}{{4{\text{x}}}}{\text{ - x + }}\dfrac{1}{{4{\text{x}}}} \\
   \Rightarrow {\text{sec}}\theta {\text{ + tan}}\theta = {\text{ 2x or }}\dfrac{1}{{{\text{2x}}}} \\
$
Hence Proved.

Note: In order to solve this type of questions the key is to have a good knowledge in trigonometric identities and relations between the tan and sec functions. Using their properties we deduce the value of tan function using the sec function.
Also, the formulae of algebraic expressions of terms of the form ${\left( {{\text{a - b}}} \right)^2}$ and ${\left( {{\text{a + b}}} \right)^2}$ are very useful to solve this problem.
Any term which is in its square form, takes both positive and negative values of itself when reduced to first order, i.e. in the term of the form, ${{\text{a}}^2}$. The solution of it can be both positive and negative values of a, i.e. +a and –a.