QUESTION

# Evaluate the following expression $\sec 70{}^\circ \sin 20{}^\circ -\cos 70{}^\circ \csc 20{}^\circ$.

Hint: First of all we can convert $\sec 70{}^\circ$ and $\csc 20{}^\circ$ in terms of $\cos \theta$ and $\sin \theta$ by using the following, $\sec \theta =\dfrac{1}{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta }$ . After that we can use $\sin \left( 90-\theta \right)=\cos \theta$ and simplify the expression further to get the answer.

Here we have been asked in the question to evaluate the given expression, $\sec 70{}^\circ \sin 20{}^\circ -\cos 70{}^\circ \csc 20{}^\circ$. Let us consider the expression that is given in the question, $E=\sec 70{}^\circ \sin 20{}^\circ -\cos 70{}^\circ \csc 20{}^\circ$.
We know that $\sec \theta =\dfrac{1}{\cos \theta }$. So, by substituting $\theta =70{}^\circ$, we get, $\sec 70{}^\circ =\dfrac{1}{\cos 70{}^\circ }$. By using this in the above expression, we get,
$E=\dfrac{\sin 20{}^\circ }{\cos 70{}^\circ }-\cos 70{}^\circ \csc 20{}^\circ$
Now, we know that $\csc \theta =\dfrac{1}{\sin \theta }$. So, by substituting $\theta =20{}^\circ$, we get, $\csc 20{}^\circ =\dfrac{1}{\sin 20{}^\circ }$. By using this in the above expression, we get,
$E=\dfrac{\sin 20{}^\circ }{\cos 70{}^\circ }-\dfrac{\cos 70{}^\circ }{\sin 20{}^\circ }$
Now, we know that $\sin \left( 90-\theta \right)=\cos \theta$. So, by substituting it, that is $\theta =70{}^\circ$, we get, $\sin \left( 90-70 \right){}^\circ =\cos 70{}^\circ \Rightarrow \sin 20{}^\circ =\cos 70{}^\circ$. By using this in the above expression, we get,
$E=\dfrac{\cos 70{}^\circ }{\cos 70{}^\circ }-\dfrac{\cos 70{}^\circ }{\cos 70{}^\circ }$
By cancelling the like terms, we get, $E=1-1$, which means that $E=0$.
Hence, we get the value of the expression given in the question, $\sec 70{}^\circ \sin 20{}^\circ -\cos 70{}^\circ \csc 20{}^\circ$ as 0.
Note: We can also simplify the given expression by using an alternate method. Let us take the expression in the first place. $E=\sec 70{}^\circ \sin 20{}^\circ -\cos 70{}^\circ \csc 20{}^\circ$. Now, we know that $\sin \left( 90-\theta \right)=\cos \theta$. So, by substituting $\theta =70{}^\circ$, we get, $\sin \left( 90-70 \right){}^\circ =\cos 70{}^\circ$. So, we get, $\sin 20{}^\circ =\cos 70{}^\circ$. We can use this in the above expression. By doing so, we get $E=\left( \sec 70{}^\circ \cos 70{}^\circ \right)-\left( \sin 20{}^\circ \csc 20{}^\circ \right)$. We also know that $\sec \theta \cos \theta =1$ and also $\csc \theta \sin \theta =1$. So, by using these in the above expression, we get, $E=1-1\Rightarrow E=0$, which is similar to the answer that we have already got.