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Evaluate the following: \[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}\].

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Hint: Transform the whole equation in terms of \[\sin \theta \]and \[\cos \theta \].

 We have to evaluate: \[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}...\left( i \right)\]
We know that \[\sin \theta =\dfrac{1}{\text{cosec}\theta }\]and \[\cos \theta =\dfrac{1}{\text{sec}\theta
 }\].
Therefore, \[\sec {{50}^{o}}=\dfrac{1}{\cos {{50}^{o}}}\]and
 \[\operatorname{cosec}{{50}^{o}}=\dfrac{1}{\sin {{50}^{o}}}\]
Now, we will put these values in equation \[\left( i \right)\].
\[=\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\operatorname{cosec}{{50}^{o}}\]
We get, \[\dfrac{\sin {{40}^{o}}}{\cos {{50}^{o}}}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}....\left( ii \right)\]

Now, we know that \[\sin \left( {{90}^{o}}-\theta \right)=\cos \theta \]
For \[\theta ={{50}^{o}}\]
We get \[\sin \left( 90-{{50}^{o}} \right)=\cos {{50}^{o}}\]
\[\Rightarrow \sin {{40}^{o}}=\cos {{50}^{o}}\]
For \[\theta ={{40}^{o}}\]
We get \[\sin \left( 90-{{40}^{o}} \right)=\cos {{40}^{o}}\]
\[\Rightarrow \sin {{50}^{o}}=\cos {{40}^{o}}\]
Putting the values of \[\sin {{40}^{o}}\]and \[\cos {{50}^{o}}\]in equation \[\left( ii \right)\].
We get, \[\dfrac{\cos {{50}^{o}}}{\cos {{50}^{o}}}+\dfrac{\cos {{40}^{o}}}{\cos {{40}^{o}}}\]
By cancelling the terms, we get
\[=1+1\]
\[=2\]
Hence, we get \[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}=2\].
Note: This question can also be easily solved by converting \[\sec \left( 90-\theta \right)\]to
 \[\operatorname{cosec}\theta \]and \[\operatorname{cosec}\left( 90-\theta \right)\]to\[\sec \theta \]as follows:
\[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}....\left( i \right)\]
We know that \[\sec \left( {{90}^{o}}-\theta \right)=\operatorname{cosec}\theta \]
For \[\theta ={{40}^{o}}\], we get \[\sec \left( {{90}^{o}}-{{40}^{o}}
 \right)=\operatorname{cosec}{{40}^{o}}\]
Therefore, \[\sec {{50}^{o}}=\operatorname{cosec}{{40}^{o}}\]
Also, we know that \[\operatorname{cosec}\left( {{90}^{o}}-\theta \right)=\sec \theta \]
For \[\theta ={{40}^{o}}\], we get \[\operatorname{cosec}\left( {{90}^{o}}-{{40}^{o}}
 \right)=sec{{40}^{o}}\]
Therefore, \[\operatorname{cosec}{{50}^{o}}=sec{{40}^{o}}\]
Now, we put values of \[\sec \left( {{50}^{o}} \right)\]and \[\operatorname{cosec}\left( {{50}^{o}}
 \right)\]in equation \[\left( i \right)\].
We get, \[\operatorname{cosec}{{40}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\sec {{40}^{o}}\]
We know that \[\operatorname{cosec}\theta .\sin \theta =1\]and \[\cos \theta .\sec \theta =1\]
By putting it in above equation,
We get, \[1+1=2\]which is our required answer.
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