Question

If ${\tan ^2}\theta = 1 - {e^2}$, then $\sec \theta + {\tan ^3}\theta \,cosec\theta$ is equal to
A) $\sqrt {\left( {2 - {e^2}} \right)}$
B) ${\left( {2 - {e^2}} \right)^{\dfrac{3}{2}}}$
C) $2 - {e^2}$
D) None of these

Answer 1

Here in this question, we have to find the value of a given trigonometric function by using a given condition. For this, first we need to rearrange the given equation by definition of trigonometric ratios and further simplify by using standard trigonometric identities and substitute the value of given condition to get the required solution.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.
Consider the question:
Given,
${\tan ^2}\theta = 1 - {e^2}$ -----(1)
We need to find the value of
$\sec \theta + {\tan ^3}\theta \,cosec\theta$
Consider,
$\Rightarrow \,\,\,\sec \theta + {\tan ^3}\theta \, \cdot \,cosec\theta$ --------(2)
By the definition of trigonometric ratios, tan is the ratio of sin and cos such as $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and the cosec is the reciprocal of sin such as $cosec\theta = \dfrac{1}{{\sin \theta }}$, then equation (2) becomes
$\Rightarrow \,\,\,\sec \theta + {\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)^3}\, \cdot \,\left( {\dfrac{1}{{\sin \theta }}} \right)$
$\Rightarrow \,\,\,\sec \theta + \left( {\dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}} \right)\, \cdot \,\left( {\dfrac{1}{{\sin \theta }}} \right)$
On simplification, we get
$\Rightarrow \,\,\,\sec \theta + \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^3}\theta }}} \right)$
$\Rightarrow \,\,\,\sec \theta + \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)\, \cdot \,\left( {\dfrac{1}{{\cos \theta }}} \right)$
Again, by the definition of trigonometric ratios, sec is the reciprocal of cos such as $sec\theta = \dfrac{1}{{\cos \theta }}$, then we have
$\Rightarrow \,\,\,\sec \theta + {\tan ^2}\theta \, \cdot \,\sec \theta$
Take $sec\theta$ as common
$\Rightarrow \,\,\,\sec \theta \left( {1 + {{\tan }^2}\theta } \right)$
As we know the standard trigonometric identity: {\text{se}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta \;$$$$ \Rightarrow \,\,\,{\text{sec}}\theta = \sqrt {1 + {{\tan }^2}\theta } \;, then the above inequality becomes
$\Rightarrow \,\,\,\sqrt {1 + {{\tan }^2}\theta } \;\left( {1 + {{\tan }^2}\theta } \right)$
$\Rightarrow \,\,\,{\left( {1 + {{\tan }^2}\theta } \right)^{\dfrac{1}{2}}} \cdot \;\left( {1 + {{\tan }^2}\theta } \right)$
We know the property of exponent ${a^m}\,{a^n} = {a^{m + n}}$, then
$\Rightarrow \,\,\,{\left( {1 + {{\tan }^2}\theta } \right)^{1 + \dfrac{1}{2}}}$
On simplification, we get
$\Rightarrow \,\,\,{\left( {1 + {{\tan }^2}\theta } \right)^{\dfrac{{2 + 1}}{2}}}$
$\Rightarrow \,\,\,{\left( {1 + {{\tan }^2}\theta } \right)^{\dfrac{3}{2}}}$ -----(3)
Given, ${\tan ^2}\theta = 1 - {e^2}$
On substituting the equation (3) becomes, then we have
$\Rightarrow \,\,\,{\left( {1 + 1 - {e^2}} \right)^{\dfrac{3}{2}}}$
On simplification we get
$\therefore \,\,\,\,\,\,\,{\left( {2 - {e^2}} \right)^{\dfrac{3}{2}}}$
Hence, it’s the required solution.
Therefore, option (B) is the correct option.

Note:
When solving the trigonometry-based questions, we have to know the definitions of all six trigonometric ratios sine, cosine, tangent, secant, cosecant and cotangent and student should know the standard three trigonometric identities and other basic trigonometric formulas on applying the formulas make the solution easily.