Question

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

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.