Question

Prove that:
$\dfrac{{(1 + \cot \theta + \tan \theta )(\sin \theta - \cos \theta )}}{{{{\sec }^3}\theta - \cos e{c^3}\theta }} = {\sin ^2}\theta {\cos ^2}\theta $

Answer 1

Hint:
By using the basic trigonometric identity given below we can prove the above expression that is $\dfrac{{(1 + \cot \theta + \tan \theta )(\sin \theta - \cos \theta )}}{{{{\sec }^3}\theta - \cos e{c^3}\theta }} = {\sin ^2}\theta {\cos ^2}\theta $ . We can prove the given statement by using ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$ . In order to solve and simplify the given expression we have to use the identity and express our given expression in the simplest form and thereby solve it.

Complete step by step solution:
$\dfrac{{(1 + \cot \theta + \tan \theta )(\sin \theta - \cos \theta )}}{{{{\sec }^3}\theta - \cos e{c^3}\theta }} = {\sin ^2}\theta {\cos ^2}\theta $
Proof:
Taking L.H.S of the given equation ,
We will get ,
$ \Rightarrow \dfrac{{(1 + \cot \theta + \tan \theta )(\sin \theta - \cos \theta )}}{{{{\sec }^3}\theta - \cos e{c^3}\theta }}$
Applying the identity ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$ ,
We will get the following,
$ \Rightarrow \dfrac{{(1 + \cot \theta + \tan \theta )(\sin \theta - \cos \theta )}}{{(\sec \theta - \cos ec \theta )({{\sec }^2}\theta + \sec \theta \cos ec \theta + \cos e{c^2}\theta )}}$
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}$ and $\cos ec \theta = \dfrac{1}{{\sin \theta }}$ ,
Now substitute these in the above equation,
We will get ,
\[ \Rightarrow \dfrac{{(1 + \cot \theta + \tan \theta )(\sin \theta - \cos \theta )}}{{\left( {\dfrac{1}{{\cos \theta }} - \dfrac{1}{{\sin \theta }}} \right)\left( {\dfrac{1}{{{{\cos }^2}\theta }} + \dfrac{1}{{\cos \theta \sin \theta }} + \dfrac{1}{{{{\sin }^2}\theta }}} \right)}}\]
\[ \Rightarrow \dfrac{{(1 + \cot \theta + \tan \theta )(\sin \theta - \cos \theta )}}{{\left( {\dfrac{{\sin \theta - \cos \theta }}{{\cos \theta \sin \theta }}} \right)\left( {\dfrac{{{{\sin }^2}\theta + \cos \theta \sin \theta + {{\cos }^2}\theta }}{{{{\cos }^2}\theta {{\sin }^2}\theta }}} \right)}}\]
We also know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , and simplify the expression,
\[ \Rightarrow \dfrac{{(1 + \cot \theta + \tan \theta )}}{{\left( {\dfrac{1}{{\cos \theta \sin \theta }}} \right)\left( {\dfrac{{1 + \cos \theta \sin \theta }}{{{{\cos }^2}\theta {{\sin }^2}\theta }}} \right)}}\]
Now, substitute $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ ,
We will get,
\[ \Rightarrow \dfrac{{\left( {1 + \dfrac{{\cos \theta }}{{\sin \theta }} + \dfrac{{\sin \theta }}{{\cos \theta }}} \right)}}{{\left( {\dfrac{1}{{\cos \theta \sin \theta }}} \right)\left( {\dfrac{{1 + \cos \theta \sin \theta }}{{{{\cos }^2}\theta {{\sin }^2}\theta }}} \right)}}\]
\[ \Rightarrow \dfrac{{\left( {\dfrac{{\cos \theta \sin \theta + {{\cos }^2}\theta + {{\sin }^2}\theta }}{{\cos \theta \sin \theta }}} \right)}}{{\left( {\dfrac{1}{{\cos \theta \sin \theta }}} \right)\left( {\dfrac{{1 + \cos \theta \sin \theta }}{{{{\cos }^2}\theta {{\sin }^2}\theta }}} \right)}}\]
Now, put ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , and simplify the expression,
\[ \Rightarrow \dfrac{{\left( {\dfrac{{\cos \theta \sin \theta + 1}}{{\cos \theta \sin \theta }}} \right)}}{{\left( {\dfrac{1}{{\cos \theta \sin \theta }}} \right)\left( {\dfrac{{1 + \cos \theta \sin \theta }}{{{{\cos }^2}\theta {{\sin }^2}\theta }}} \right)}}\]
\[
   \Rightarrow \dfrac{1}{{\left( {\dfrac{1}{{{{\cos }^2}\theta {{\sin }^2}\theta }}} \right)}} \\
   = {\cos ^2}\theta {\sin ^2}\theta \\
 \]
It is equal to R.H.S.
Since L.H.S is equal to R.H.S.
Hence proved.

Note:
Some other equations needed for solving these types of problem are:
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$,
$\sec \theta = \dfrac{1}{{\cos \theta }}$ , $\cos ec \theta = \dfrac{1}{{\sin \theta }}$ , $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$

In order to solve and simplify the given expression we have to use the identity and express our given expression in the simplest form and thereby solve it. Also, while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.