
Find the value of \[\dfrac{{\left( {\tan {{80}^ \circ } - \tan {{10}^ \circ }} \right)}}{{\tan {{70}^ \circ }}}\].
A. 0
B. 1
C. 2
D. 3
Answer
232.5k+ views
Hint: In this question, we have to determine the value of the trigonometric expression \[\dfrac{{\left( {\tan {{80}^ \circ } - \tan {{10}^ \circ }} \right)}}{{\tan {{70}^ \circ }}}\]. For this, we have to consider an equation \[{80^ \circ } - {10^ \circ } = {70^ \circ }\]. So, we need to take tan on both sides. After simplification of it and using the formula of \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\], we get the desired result.
Formula used: We will use the following trigonometric identities for solving this example.
\[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]
Complete step-by-step answer:
Consider \[{80^ \circ } - {10^ \circ } = {70^ \circ }\]
So, by taking ‘tan’ on both sides, we get
\[\tan \left( {{{80}^ \circ } - {{10}^ \circ }} \right) = \tan {70^ \circ }\]
But \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]
Thus, by applying the above trigonometric identity, we get
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{1 + \tan {{80}^ \circ }\tan {{10}^ \circ }}} = \tan {70^ \circ }\]
By simplifying further, we get
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 1 + \tan {80^ \circ }\tan {10^ \circ }\]
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 1 + \tan {80^ \circ }\left( {\tan ({{90}^ \circ } - {80}^ \circ }) \right)\]
But \[\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \]
So, by applying the above trigonometric identity, we get
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 1 + \tan {80^ \circ }\cot {80^ \circ }\]
We know that \[\tan A\cot A = 1\]
Hence, we get
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 1 + 1\]
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 2\]
Hence, the value of \[\dfrac{{\left( {\tan {{80}^ \circ } - \tan {{10}^ \circ }} \right)}}{{\tan {{70}^ \circ }}}\] is \[2\].
Therefore, the correct option is (C).
Note: Many students make mistakes in the simplification part. They get confused in applying the right trigonometric identity in the appropriate place since there are so many identities there. This problem can be also solved by converting each tan function into a sine and cos functions .
Formula used: We will use the following trigonometric identities for solving this example.
\[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]
Complete step-by-step answer:
Consider \[{80^ \circ } - {10^ \circ } = {70^ \circ }\]
So, by taking ‘tan’ on both sides, we get
\[\tan \left( {{{80}^ \circ } - {{10}^ \circ }} \right) = \tan {70^ \circ }\]
But \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]
Thus, by applying the above trigonometric identity, we get
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{1 + \tan {{80}^ \circ }\tan {{10}^ \circ }}} = \tan {70^ \circ }\]
By simplifying further, we get
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 1 + \tan {80^ \circ }\tan {10^ \circ }\]
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 1 + \tan {80^ \circ }\left( {\tan ({{90}^ \circ } - {80}^ \circ }) \right)\]
But \[\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \]
So, by applying the above trigonometric identity, we get
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 1 + \tan {80^ \circ }\cot {80^ \circ }\]
We know that \[\tan A\cot A = 1\]
Hence, we get
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 1 + 1\]
\[\dfrac{{\tan {{80}^ \circ } - \tan {{10}^ \circ }}}{{\tan {{70}^ \circ }}} = 2\]
Hence, the value of \[\dfrac{{\left( {\tan {{80}^ \circ } - \tan {{10}^ \circ }} \right)}}{{\tan {{70}^ \circ }}}\] is \[2\].
Therefore, the correct option is (C).
Note: Many students make mistakes in the simplification part. They get confused in applying the right trigonometric identity in the appropriate place since there are so many identities there. This problem can be also solved by converting each tan function into a sine and cos functions .
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