Question

If we have a trigonometric expression ${\text{cosec}}\theta + \cot \theta = {\text{P}}$ then the value of $\cos \theta $ is

Answer 1

Hint: In this particular question use the concept that $\cos ec x = \dfrac{1}{{\sin \theta }}$ and $\cot x = \dfrac{{\cos \theta }}{{\sin \theta }}$ so use these properties and try to simplify the given trigonometric equation by squaring on both sides and later on use standard trigonometric identity such as ${\sin ^2}\theta + {\cos ^2}\theta = 1$ to reach the solution of the question.

Complete step-by-step solution:
Given trigonometric equation is
${\text{cosec}}\theta + \cot \theta = {\text{P}}$
Now as we know that $\cos ec x =\dfrac{1}{{\sin \theta }}$ and $\cot x = \dfrac{{\cos \theta }}{{\sin \theta }}$, so use these properties in the above equation we have,
$ \Rightarrow \dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }} = {\text{P}}$
$ \Rightarrow \dfrac{{1 + \cos \theta }}{{\sin \theta }} = {\text{P}}$
Now squaring on both sides we have,
$ \Rightarrow {\left( {\dfrac{{1 + \cos \theta }}{{\sin \theta }}} \right)^2} = {{\text{P}}^2}$
\[ \Rightarrow \left( {\dfrac{{{{\left( {1 + \cos \theta } \right)}^2}}}{{{{\sin }^2}\theta }}} \right) = {{\text{P}}^2}\]
Now as we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$ so, ${\sin ^2}\theta = 1 - {\cos ^2}\theta $so use this property in the above equation we have,
\[ \Rightarrow \dfrac{{{{\left( {1 + \cos \theta } \right)}^2}}}{{1 - {{\cos }^2}\theta }} = {{\text{P}}^2}\]
Now as we know that ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ so we have,
\[ \Rightarrow \dfrac{{{{\left( {1 + \cos \theta } \right)}^2}}}{{\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)}} = {{\text{P}}^2}\]
\[ \Rightarrow \dfrac{{\left( {1 + \cos \theta } \right)}}{{\left( {1 - \cos \theta } \right)}} = {{\text{P}}^2}\]
Now simplify it we have,
\[ \Rightarrow \left( {1 + \cos \theta } \right) = {{\text{P}}^2}\left( {1 - \cos \theta } \right)\]
\[ \Rightarrow \left( {1 + \cos \theta } \right) = {{\text{P}}^2} - {{\text{P}}^2}\cos \theta \]
\[ \Rightarrow {{\text{P}}^2}\cos \theta + \cos \theta = {{\text{P}}^2} - 1\]
\[ \Rightarrow \cos \theta \left( {{{\text{P}}^2} + 1} \right) = {{\text{P}}^2} - 1\]
\[ \Rightarrow \cos \theta = \dfrac{{{{\text{P}}^2} - 1}}{{{{\text{P}}^2} + 1}}\]
So this is the required value of the $\cos \theta $
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic standard trigonometric properties as well as identities which is all stated above, and always recall the common known fact that ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, so apply this as above then simplify we will get the required answer.