Question

Show that \[\cos \theta \left[ {\dfrac{1}{{1 - \sin \theta }} - \dfrac{1}{{1 + \sin \theta }}} \right]\] can be written in the form \[k\]. \[\tan \theta \]. Find the value \[k\].

Answer 1

Hint: To solve this question first we assume a variable equal to the given expression. Then we equate that to the multiplication of constant and tangent functions. Then we simplify that given expression by taking LCM at the denominator and simplify that expression and convert that to tangent function and the constant term is the value of k.

Complete step by step answer:
Given that:
\[\cos \theta \left[ {\dfrac{1}{{1 - \sin \theta }} - \dfrac{1}{{1 + \sin \theta }}} \right]\]
To find:
Express the given trigonometric expression in terms of \[\tan \theta \] and find the coefficient of \[\tan \theta \].
Let the given expression be equal to the variable \[x\].
\[x = \cos \theta \left[ {\dfrac{1}{{1 - \sin \theta }} - \dfrac{1}{{1 + \sin \theta }}} \right]\]
Now on taking the LCM in the denominator.
\[x = \cos \theta \left[ {\dfrac{{1 + \sin \theta - \left( {1 - \sin \theta } \right)}}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}}} \right]\]
Further simplifying the numerator and denominator.
\[x = \cos \theta \left[ {\dfrac{{1 + \sin \theta - 1 + \sin \theta }}{{1 - {{\sin }^2}\theta }}} \right]\]
On simplifying the equation.
\[x = \cos \theta \left[ {\dfrac{{2\sin \theta }}{{1 - {{\sin }^2}\theta }}} \right]\]
On using the identity of trigonometry \[1 - {\sin ^2}\theta = {\cos ^2}\theta \]
\[x = \cos \theta \left[ {\dfrac{{2\sin \theta }}{{{{\cos }^2}\theta }}} \right]\]
On canceling the common factor from numerator and denominator.
\[x = \dfrac{{2\sin \theta }}{{\cos \theta }}\]
On using the trigonometry relation \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[x = 2\tan \theta \]
Here we have to make the given expression in terms of a tan trigonometric function and a constant with that in an equation that we find that constant.
\[\cos \theta \left[ {\dfrac{1}{{1 - \sin \theta }} - \dfrac{1}{{1 + \sin \theta }}} \right] = k\tan \theta \]
On putting the value from the simplified version.
\[2\tan \theta = k\tan \theta \]
On comparing both sides of the equation the value of \[k\] is:
\[k = 2\]
The value of k is 2.

Note:
Students must know all the relations of trigonometry, formulas of trigonometry, and identities of trigonometry. If the more simplified version is given in options then we simplify that according to the options and mark the correct answer.