Question

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

Answer 1

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.