Question

Simplify the trigonometry expression \[\left[ {\dfrac{{\sin \theta }}{{\left( {1 - \cot \theta } \right)}}} \right] + \left[ {\dfrac{{\cos \theta }}{{\left( {1 - \tan \theta } \right)}}} \right]\].
A. 0
B. 1
C. \[\cos \theta - \sin \theta \]
D. \[\cos \theta + \sin \theta \]

Answer 1

Hint: First we will rewrite \[\cot \theta \] and \[\tan \theta \] in terms of \[\sin \theta \] and \[\cos \theta \]. Then simplify each term. Then add or subtract both terms. Apply the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] in the numerator and cancel out common term to get desired result.

Formula Used:
\[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Complete step by step solution:
Given expression is \[\left[ {\dfrac{{\sin \theta }}{{\left( {1 - \cot \theta } \right)}}} \right] + \left[ {\dfrac{{\cos \theta }}{{\left( {1 - \tan \theta } \right)}}} \right]\]
Now apply the formula \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\] and \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[ = \left[ {\dfrac{{\sin \theta }}{{\left( {1 - \dfrac{{\cos \theta }}{{\sin \theta }}} \right)}}} \right] + \left[ {\dfrac{{\cos \theta }}{{\left( {1 - \dfrac{{\sin \theta }}{{\cos \theta }}} \right)}}} \right]\]
Simplify the denominator of each term
\[ = \left[ {\dfrac{{\sin \theta }}{{\left( {\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta }}} \right)}}} \right] + \left[ {\dfrac{{\cos \theta }}{{\left( {\dfrac{{\cos \theta - \sin \theta }}{{\cos \theta }}} \right)}}} \right]\]
\[ = \left[ {\dfrac{{{{\sin }^2}\theta }}{{\sin \theta - \cos \theta }}} \right] + \left[ {\dfrac{{{{\cos }^2}\theta }}{{\cos \theta - \sin \theta }}} \right]\]
Take common -1 from the denominator of the second term
\[ = \left[ {\dfrac{{{{\sin }^2}\theta }}{{\sin \theta - \cos \theta }}} \right] - \left[ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta - \cos \theta }}} \right]\]
Now subtract both terms
\[ = \left[ {\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{\sin \theta - \cos \theta }}} \right]\]
Apply the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
\[ = \dfrac{{\left( {\sin \theta - \cos \theta } \right)\left( {\sin \theta + \cos \theta } \right)}}{{\sin \theta - \cos \theta }}\]
Cancel out \[\left( {\sin \theta - \cos \theta } \right)\]
\[ = \cos \theta + \sin \theta \]

Option D is the correct option.

Note: Students often confused with the formula \[{\sin ^2}\theta - {\cos ^2}\theta = 1\] and \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]. They put \[{\sin ^2}\theta - {\cos ^2}\theta = 1\] in the step \[\left[ {\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{\sin \theta - \cos \theta }}} \right]\] and get \[\dfrac{1}{{\sin \theta - \cos \theta }}\] which is a wrong result. The correct formula is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and in this step \[\left[ {\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{\sin \theta - \cos \theta }}} \right]\] we will apply \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] to get desired result.