Question

How do you prove $\dfrac{\sin x+\cos x}{\sin x}-\dfrac{\cos x-\sin x}{\cos x}=\sec x\cos ecx?$

Answer 1

Hint: We have to make use of basic mathematics. We are going to take L.C.M. And we are also going to use some trigonometric identities. There is more than one way to do sums in trigonometry. Anyway we get the same answer. We are going to make use of ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, $\dfrac{1}{\cos x}=\sec x$, and$\dfrac{1}{\sin x}=\cos ecx$ .

Complete step-by-step solution:
We shall first solve the left-hand side of the question.
First of all, let us take L.C.M i.e cross-multiply the denominators to each other’s numerator. It goes like this :
\[\begin{align}
  & \Rightarrow \dfrac{\sin x+\cos x}{\sin x}-\dfrac{\cos x-\sin x}{\cos x} \\
 & \Rightarrow \dfrac{\left( \cos x \right)\left( \cos x+\sin x \right)}{\sin x\cos x}-\dfrac{\left( \sin x \right)\left( \cos x-\sin x \right)}{\sin x\cos x} \\
 & \Rightarrow \dfrac{{{\cos }^{2}}x+\cos x\sin x-\sin x\cos x+{{\sin }^{2}}x}{\sin x\cos x} \\
\end{align}\]
We can see that there are like terms with the opposite signs in the third step and they cancel out.
So, \[\cos x\sin x-\sin x\cos x\] will cancel out each other.
Now it looks like the following :
\[\Rightarrow \dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\sin x\cos x}\]
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . This is one of the most basic identities that we have of trigonometry. Upon applying it in the above equation, we get the following :
\[\Rightarrow \dfrac{1}{\sin x\cos x}\]
We also know that $\dfrac{1}{\cos x}=\sec x$ and $\dfrac{1}{\sin x}=\cos ecx$ . Upon applying it in the above equation we get the following :
\[\begin{align}
  & \Rightarrow \dfrac{1}{\sin x\cos x} \\
 & \Rightarrow \sec x\cos ecx \\
\end{align}\]
This is the result of the left-hand side of our question after simplifying.
Now let us see what we got on our right-hand side.
We got \[\sec x\cos ecx\] .
So the left-hand side is equal to the right-hand side of the question.
Hence proved that $\dfrac{\sin x+\cos x}{\sin x}-\dfrac{\cos x-\sin x}{\cos x}=\sec x\cos ecx$ .

Note: We should thoroughly remember all the trigonometric identities, functions, values of each trigonometric function, every trigonometric function’s domain, and their range. We can easily solve the questions if all these are at our fingertips.