
What is the value of \[\cot \left[ {{{\cos }^{ - 1}}\left( {\dfrac{7}{{25}}} \right)} \right]\]?
A. \[\dfrac{{25}}{{24}}\]
B. \[\dfrac{{25}}{7}\]
C. \[\dfrac{{7}}{{24}}\]
D. None of these
Answer
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Hint: First we will assume \[{\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right) = \theta \]. Then by using trigonometry identity we will calculate \[\sin \theta \]. Then we will calculate \[\cot \theta \] and from this, we will calculate \[\theta \]. After that, we will substitute the value of \[\theta \] the given expression and simplify it.
Formula used:
Trigonometry identity: \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Trigonometry ratio: \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
Inverse function: \[\cot \left( {{{\cot }^{ - 1}}\theta } \right) = \theta \]
Complete step by step solution:
Given expression is \[\cot \left[ {{{\cos }^{ - 1}}\left( {\dfrac{7}{{25}}} \right)} \right]\]
Assume that \[{\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right) = \theta \].
Taking \[\cos \] function both sides
\[\cos \left[ {{{\cos }^{ - 1}}\left( {\dfrac{7}{{25}}} \right)} \right] = \cos \theta \]
Apply inverse function
\[\cos \theta = \dfrac{7}{{25}}\] [Since \[\cos \left[ {{{\cos }^{ - 1}}\theta } \right] = \theta \]]
Now applying trigonometry identity to calculate \[\sin \theta \]
\[{\sin ^2}\theta + {\left( {\dfrac{7}{{25}}} \right)^2} = 1\] [Since \[\cos \theta = \dfrac{7}{{25}}\]]
\[ \Rightarrow {\sin ^2}\theta + \dfrac{{49}}{{625}} = 1\]
\[ \Rightarrow {\sin ^2}\theta = 1 - \dfrac{{49}}{{625}}\]
\[ \Rightarrow {\sin ^2}\theta = \dfrac{{576}}{{625}}\]
Taking square root both sides
\[ \Rightarrow \sin \theta = \sqrt {\dfrac{{576}}{{625}}} \] [Taking positive value since \[\cos \theta \] is positive]
\[ \Rightarrow \sin \theta = \dfrac{{24}}{{25}}\]
Now we will calculate \[\cot \theta \].
\[\cot \theta = \dfrac{{\dfrac{7}{{25}}}}{{\dfrac{{24}}{{25}}}}\]
\[ \Rightarrow \cot \theta = \dfrac{7}{{24}}\]
Taking \[{\cot ^{ - 1}}\] on both sides
\[ \Rightarrow {\cot ^{ - 1}}\left( {\cot \theta } \right) = {\cot ^{ - 1}}\left( {\dfrac{7}{{24}}} \right)\]
\[ \Rightarrow \theta = {\cot ^{ - 1}}\left( {\dfrac{7}{{24}}} \right)\]
Therefore \[\theta = {\cot ^{ - 1}}\left( {\dfrac{7}{{24}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right)\]
Now substitute \[{\cot ^{ - 1}}\left( {\dfrac{7}{{24}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right)\] in the given expression
\[\cot \left[ {{{\cos }^{ - 1}}\left( {\dfrac{7}{{25}}} \right)} \right] = \cot \left[ {{{\cot }^{ - 1}}\left( {\dfrac{7}{{24}}} \right)} \right]\]
Apply inverse function
\[ = \dfrac{7}{{24}}\]
Hence option C is the correct option.
Additional information:
In the first quadrant, all trigonometry ratios are positive. In the second quadrant, only \[\sin\] and \[\csc\] are positive and the rest trigonometry ratios are negative. In the third quadrant, only \[\tan\] and \[\cot\] are positive and the rest trigonometry ratios are negative. In the fourth quadrant, only \[\cos\] and \[\sec\] are positive and the rest trigonometry ratios are negative.
Note: Many students often take negative value of \[\sin \theta = - \sqrt {\dfrac{{576}}{{625}}} \]. Since \[\cos \theta = \dfrac{7}{{25}}\] is positive value so \[\theta \] must be lie in the first quadrant. In the first quadrant, all trigonometry ratios are positive. So the value of \[\sin \theta \] must be a positive value.
Formula used:
Trigonometry identity: \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Trigonometry ratio: \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
Inverse function: \[\cot \left( {{{\cot }^{ - 1}}\theta } \right) = \theta \]
Complete step by step solution:
Given expression is \[\cot \left[ {{{\cos }^{ - 1}}\left( {\dfrac{7}{{25}}} \right)} \right]\]
Assume that \[{\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right) = \theta \].
Taking \[\cos \] function both sides
\[\cos \left[ {{{\cos }^{ - 1}}\left( {\dfrac{7}{{25}}} \right)} \right] = \cos \theta \]
Apply inverse function
\[\cos \theta = \dfrac{7}{{25}}\] [Since \[\cos \left[ {{{\cos }^{ - 1}}\theta } \right] = \theta \]]
Now applying trigonometry identity to calculate \[\sin \theta \]
\[{\sin ^2}\theta + {\left( {\dfrac{7}{{25}}} \right)^2} = 1\] [Since \[\cos \theta = \dfrac{7}{{25}}\]]
\[ \Rightarrow {\sin ^2}\theta + \dfrac{{49}}{{625}} = 1\]
\[ \Rightarrow {\sin ^2}\theta = 1 - \dfrac{{49}}{{625}}\]
\[ \Rightarrow {\sin ^2}\theta = \dfrac{{576}}{{625}}\]
Taking square root both sides
\[ \Rightarrow \sin \theta = \sqrt {\dfrac{{576}}{{625}}} \] [Taking positive value since \[\cos \theta \] is positive]
\[ \Rightarrow \sin \theta = \dfrac{{24}}{{25}}\]
Now we will calculate \[\cot \theta \].
\[\cot \theta = \dfrac{{\dfrac{7}{{25}}}}{{\dfrac{{24}}{{25}}}}\]
\[ \Rightarrow \cot \theta = \dfrac{7}{{24}}\]
Taking \[{\cot ^{ - 1}}\] on both sides
\[ \Rightarrow {\cot ^{ - 1}}\left( {\cot \theta } \right) = {\cot ^{ - 1}}\left( {\dfrac{7}{{24}}} \right)\]
\[ \Rightarrow \theta = {\cot ^{ - 1}}\left( {\dfrac{7}{{24}}} \right)\]
Therefore \[\theta = {\cot ^{ - 1}}\left( {\dfrac{7}{{24}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right)\]
Now substitute \[{\cot ^{ - 1}}\left( {\dfrac{7}{{24}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right)\] in the given expression
\[\cot \left[ {{{\cos }^{ - 1}}\left( {\dfrac{7}{{25}}} \right)} \right] = \cot \left[ {{{\cot }^{ - 1}}\left( {\dfrac{7}{{24}}} \right)} \right]\]
Apply inverse function
\[ = \dfrac{7}{{24}}\]
Hence option C is the correct option.
Additional information:
In the first quadrant, all trigonometry ratios are positive. In the second quadrant, only \[\sin\] and \[\csc\] are positive and the rest trigonometry ratios are negative. In the third quadrant, only \[\tan\] and \[\cot\] are positive and the rest trigonometry ratios are negative. In the fourth quadrant, only \[\cos\] and \[\sec\] are positive and the rest trigonometry ratios are negative.
Note: Many students often take negative value of \[\sin \theta = - \sqrt {\dfrac{{576}}{{625}}} \]. Since \[\cos \theta = \dfrac{7}{{25}}\] is positive value so \[\theta \] must be lie in the first quadrant. In the first quadrant, all trigonometry ratios are positive. So the value of \[\sin \theta \] must be a positive value.
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