Question

If \[\dfrac{{ - \pi }}{2} < \theta < \dfrac{\pi }{2}\]and \[\theta \ne \dfrac{\pi }{4}\], then what is the value of \[\cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right)\]
A.0
B.-1
C.1
D.-2
E.2

Answer 1

Hint: First we will concert \[\cot \] to \[\tan \]by applying the formula \[\cot \theta = \dfrac{1}{{\tan \theta }}\]. After that we will apply the formula \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]. Then we will put \[\tan \dfrac{\pi }{4} = 1\]. Then simplify the rational function.

Formula Used:

We will use trigonometric identity
\[\cot \theta = \dfrac{1}{{\tan \theta }}\], \[\tan \dfrac{\pi }{4} = 1\] and angle sum identities of tan i.e., \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and
 angle difference identities i.e., \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\].

Complete step by step solution:

Given \[\cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right)\]
Now we will convert cot to tan by using trigonometric identity, \[\cot \theta = \dfrac{1}{{\tan \theta }}\], we will get,
\[ \Rightarrow \cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right) = \left( {\dfrac{1}{{\tan \left( {\dfrac{\pi }{4} + \theta } \right)}}} \right)\left( {\dfrac{1}{{\tan \left( {\dfrac{\pi }{4} - \theta } \right)}}} \right)\]
Now we will use the angle sum identities of tan i.e., \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and angle difference identities i.e., \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\], we will get,
\[ \Rightarrow \cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right) = \left( {\dfrac{1}{{\dfrac{{\tan \dfrac{\pi }{4} + \tan \theta }}{{1 - \tan \dfrac{\pi }{4}\tan \theta }}}}} \right)\left( {\dfrac{1}{{\dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{1 + \tan \dfrac{\pi }{4}\tan \theta }}}}} \right)\].
Now we will use the fact that\[\tan \dfrac{\pi }{4} = 1\], and we will substitute the value in the expression, we will get,
\[ \Rightarrow \cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right) = \left( {\dfrac{1}{{\dfrac{{1 + \tan \theta }}{{1 - 1\left( {\tan \theta } \right)}}}}} \right)\left( {\dfrac{1}{{\dfrac{{1 - \tan \theta }}{{1 + 1\left( {\tan \theta } \right)}}}}} \right)\]
Now we will simplify the expression we will get,
\[ \Rightarrow \cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right) = \left( {\dfrac{1}{{\dfrac{{1 + \tan \theta }}{{1 - \tan \theta }}}}} \right)\left( {\dfrac{1}{{\dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}}}} \right)\]
Now we will further simplify the expression we will get,
\[ \Rightarrow \cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right) = \left( {\dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)\left( {\dfrac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right)\]
Now we will eliminate the like terms we will get,
\[ \Rightarrow \cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right) = 1\]
The correct option is C.
Note: students are confused with the identity formula of \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]. Remember the sign between \[\tan A\] and \[\tan B\]of the numerator is same as the sign between \[A\] and \[B\] and the sign between \[1\] and \[\tan A\tan B\] must opposite sign to the sign between \[\tan A\] and \[\tan B\].