# \[\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}=?\]

Last updated date: 28th Mar 2023

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Answer

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Hint: First of all convert the given expression in terms of \[\sin \theta \] and \[\cos \theta \] by using \[\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta }\] and \[\text{cosec}\theta =\dfrac{1}{\sin \theta }\]. Then use \[\sin \left( 90-\theta \right)=\cos \theta \] and \[\cos \left( 90-\theta \right)=\sin \theta \] and substitute \[\theta ={{50}^{0}}\text{ and 4}{{\text{0}}^{o}}\] to get the final value of the expression.

Complete step-by-step answer:

Here, we have to find the value of the expression, \[\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\text{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\text{cosec5}{{\text{0}}^{o}}\].

Let us consider the expression given in the question

\[E=\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}\]

We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\text{ and }\cot \theta =\dfrac{\cos \theta }{\sin \theta }\].

By applying these in the above expression, we get,

\[E=\dfrac{\dfrac{\sin {{50}^{o}}}{\cos {{50}^{o}}}+\sec {{50}^{o}}}{\dfrac{\cos {{40}^{o}}}{\sin {{40}^{o}}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}\]

We also know that \[\sec \theta =\dfrac{1}{\cos \theta }\] and \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]. By applying these in the above expression, we get,

\[E=\dfrac{\dfrac{\sin {{50}^{o}}}{\cos {{50}^{o}}}+\dfrac{1}{\cos {{50}^{o}}}}{\dfrac{\cos {{40}^{o}}}{\sin {{40}^{o}}}+\dfrac{1}{\sin {{40}^{o}}}}+\cos {{40}^{o}}.\dfrac{1}{\sin {{50}^{o}}}\]

By simplifying the above expression, we get

\[E=\dfrac{\dfrac{\left( \sin {{50}^{o}}+1 \right)}{\cos {{50}^{o}}}}{\dfrac{\left( \cos {{40}^{o}}+1 \right)}{\sin {{40}^{o}}}}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}\]

We know that, \[\sin \left( 90-\theta \right)=\cos \theta \]

By substituting, \[\theta ={{50}^{o}}\], we get,

\[\sin \left( 90-{{50}^{o}} \right)=\cos {{50}^{o}}\]

Or, \[\sin \left( {{40}^{o}} \right)=\cos \left( {{50}^{o}} \right)\]

By substituting \[\sin {{40}^{o}}=\cos {{50}^{o}}\] in the above expression, we get,

\[E=\dfrac{\dfrac{\left( \sin {{50}^{o}}+1 \right)}{\cos {{50}^{o}}}}{\dfrac{\left( \cos {{40}^{o}}+1 \right)}{\cos {{50}^{o}}}}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}\]

By cancelling the like terms, we get,

\[E=\dfrac{\left( \sin {{50}^{o}} \right)+1}{\left( \cos {{40}^{o}} \right)+1}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}\]

We also know that \[\cos \left( {{90}^{o}}-\theta \right)=\sin \theta \]

By substituting \[\theta ={{50}^{o}}\], we get,

\[\cos \left( {{90}^{o}}-{{50}^{o}} \right)=\sin {{50}^{o}}\]

Or, \[\cos \left( {{40}^{o}} \right)=\sin \left( {{50}^{o}} \right)\]

By substituting \[\cos \left( {{40}^{o}} \right)=\sin \left( {{50}^{o}} \right)\] in the above expression, we get,

\[E=\dfrac{\left( \sin {{50}^{o}} \right)+1}{\left( \sin {{50}^{o}} \right)+1}+\dfrac{\sin {{50}^{o}}}{\sin {{50}^{o}}}\]

By cancelling the like terms, we get,

\[E=\dfrac{1}{1}+\dfrac{1}{1}\]

Or, \[E=1+1=2\]

Hence, the value of the expression \[\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}\] is equal to 2.

Note: Students can also solve this question directly in this way.

Let the expression be

\[E=\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}\]

We know that \[\tan \left( {{90}^{o}}-\theta \right)=\cot \theta \] and \[\sec \left( {{90}^{o}}-\theta \right)=\operatorname{cosec}\theta \]

By substituting \[\theta ={{40}^{o}}\], we get,

\[\tan {{50}^{o}}=\cot {{40}^{o}}\] and \[\sec {{50}^{o}}=\operatorname{cosec}{{40}^{o}}\].

By substituting the value of \[\tan {{50}^{o}}\] and \[\sec {{50}^{o}}\] in the above expression, we get,

\[E=\dfrac{\left( \cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}} \right)}{\left( \cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}} \right)}+\cos {{40}^{o}}\operatorname{cosec}{{40}^{o}}\]

By cancelling the like terms, we get,

\[E=1+\cos {{40}^{o}}\operatorname{cosec}{{40}^{o}}\]

We know that, \[\operatorname{cosec}\left( 90-\theta \right)=\sec \theta \]

By substituting \[\theta ={{50}^{o}}\], we get,

\[\operatorname{cosec}\left( {{40}^{o}} \right)=\sec {{50}^{o}}\]

By substituting the value of \[\operatorname{cosec}{{40}^{o}}\] in the above expression, we get

\[E=1+\cos {{40}^{o}},\sec {{40}^{o}}\]

We know that \[\cos \theta .\sec \theta =1\]. By applying this in the above expression, we get

\[E=1+1=2\]

Hence, the value of the given expression is 2.

Complete step-by-step answer:

Here, we have to find the value of the expression, \[\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\text{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\text{cosec5}{{\text{0}}^{o}}\].

Let us consider the expression given in the question

\[E=\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}\]

We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\text{ and }\cot \theta =\dfrac{\cos \theta }{\sin \theta }\].

By applying these in the above expression, we get,

\[E=\dfrac{\dfrac{\sin {{50}^{o}}}{\cos {{50}^{o}}}+\sec {{50}^{o}}}{\dfrac{\cos {{40}^{o}}}{\sin {{40}^{o}}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}\]

We also know that \[\sec \theta =\dfrac{1}{\cos \theta }\] and \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]. By applying these in the above expression, we get,

\[E=\dfrac{\dfrac{\sin {{50}^{o}}}{\cos {{50}^{o}}}+\dfrac{1}{\cos {{50}^{o}}}}{\dfrac{\cos {{40}^{o}}}{\sin {{40}^{o}}}+\dfrac{1}{\sin {{40}^{o}}}}+\cos {{40}^{o}}.\dfrac{1}{\sin {{50}^{o}}}\]

By simplifying the above expression, we get

\[E=\dfrac{\dfrac{\left( \sin {{50}^{o}}+1 \right)}{\cos {{50}^{o}}}}{\dfrac{\left( \cos {{40}^{o}}+1 \right)}{\sin {{40}^{o}}}}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}\]

We know that, \[\sin \left( 90-\theta \right)=\cos \theta \]

By substituting, \[\theta ={{50}^{o}}\], we get,

\[\sin \left( 90-{{50}^{o}} \right)=\cos {{50}^{o}}\]

Or, \[\sin \left( {{40}^{o}} \right)=\cos \left( {{50}^{o}} \right)\]

By substituting \[\sin {{40}^{o}}=\cos {{50}^{o}}\] in the above expression, we get,

\[E=\dfrac{\dfrac{\left( \sin {{50}^{o}}+1 \right)}{\cos {{50}^{o}}}}{\dfrac{\left( \cos {{40}^{o}}+1 \right)}{\cos {{50}^{o}}}}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}\]

By cancelling the like terms, we get,

\[E=\dfrac{\left( \sin {{50}^{o}} \right)+1}{\left( \cos {{40}^{o}} \right)+1}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}\]

We also know that \[\cos \left( {{90}^{o}}-\theta \right)=\sin \theta \]

By substituting \[\theta ={{50}^{o}}\], we get,

\[\cos \left( {{90}^{o}}-{{50}^{o}} \right)=\sin {{50}^{o}}\]

Or, \[\cos \left( {{40}^{o}} \right)=\sin \left( {{50}^{o}} \right)\]

By substituting \[\cos \left( {{40}^{o}} \right)=\sin \left( {{50}^{o}} \right)\] in the above expression, we get,

\[E=\dfrac{\left( \sin {{50}^{o}} \right)+1}{\left( \sin {{50}^{o}} \right)+1}+\dfrac{\sin {{50}^{o}}}{\sin {{50}^{o}}}\]

By cancelling the like terms, we get,

\[E=\dfrac{1}{1}+\dfrac{1}{1}\]

Or, \[E=1+1=2\]

Hence, the value of the expression \[\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}\] is equal to 2.

Note: Students can also solve this question directly in this way.

Let the expression be

\[E=\dfrac{\tan {{50}^{o}}+\sec {{50}^{o}}}{\cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}}}+\cos {{40}^{o}}.\operatorname{cosec}{{50}^{o}}\]

We know that \[\tan \left( {{90}^{o}}-\theta \right)=\cot \theta \] and \[\sec \left( {{90}^{o}}-\theta \right)=\operatorname{cosec}\theta \]

By substituting \[\theta ={{40}^{o}}\], we get,

\[\tan {{50}^{o}}=\cot {{40}^{o}}\] and \[\sec {{50}^{o}}=\operatorname{cosec}{{40}^{o}}\].

By substituting the value of \[\tan {{50}^{o}}\] and \[\sec {{50}^{o}}\] in the above expression, we get,

\[E=\dfrac{\left( \cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}} \right)}{\left( \cot {{40}^{o}}+\operatorname{cosec}{{40}^{o}} \right)}+\cos {{40}^{o}}\operatorname{cosec}{{40}^{o}}\]

By cancelling the like terms, we get,

\[E=1+\cos {{40}^{o}}\operatorname{cosec}{{40}^{o}}\]

We know that, \[\operatorname{cosec}\left( 90-\theta \right)=\sec \theta \]

By substituting \[\theta ={{50}^{o}}\], we get,

\[\operatorname{cosec}\left( {{40}^{o}} \right)=\sec {{50}^{o}}\]

By substituting the value of \[\operatorname{cosec}{{40}^{o}}\] in the above expression, we get

\[E=1+\cos {{40}^{o}},\sec {{40}^{o}}\]

We know that \[\cos \theta .\sec \theta =1\]. By applying this in the above expression, we get

\[E=1+1=2\]

Hence, the value of the given expression is 2.

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