Question

How do you simplify this expression \[1+\dfrac{\tan x}{\sin x}+\cos x\] ?

Answer 1

Hint: The above mentioned problem is a simple demonstration of trigonometric equations and expressions. Before solving this type of questions which involve simplifying, we need to recall some of the basic equations and relations of trigonometry. We start off the solution by converting the given problem into simpler forms which have \[\sin x\] and \[\cos x\] as the main terms. The equations are described as:
\[\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
 & \cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
We also need to remember some of the basic defined relations, which are:
\[\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
 & \text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1 \\
\end{align}\]

Complete step by step answer:
Now starting off with the solution, we can firstly notice the form given is \[\tan x\] , and on rewriting this value of \[\tan x\] as \[\tan x=\dfrac{\sin x}{\cos x}\] , we proceed for the problem as,
\[\Rightarrow 1+\dfrac{\dfrac{\sin x}{\cos x}}{\sin x}+\cos x\]
Rewriting in the simpler form we get,
\[\Rightarrow 1+\dfrac{\sin x}{\sin x\cos x}+\cos x\]
Now cancelling off the \[\sin x\] terms, we get,
\[\Rightarrow 1+\dfrac{1}{\cos x}+\cos x\]
Now, we can write \[\sec x=\dfrac{1}{\cos x}\] , thus replacing this in the original equation, we get,
\[\Rightarrow 1+\sec x+\cos x\]
We now can remember that,
\[\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}\]
Using the relations of trigonometry, we can simplify it further by writing,
\[\cos x={{\cos }^{2}}\dfrac{x}{2}-\left( 1-{{\cos }^{2}}\dfrac{x}{2} \right)\]
Now, on adding the like terms we get,
\[\Rightarrow \cos x=2{{\cos }^{2}}\dfrac{x}{2}-1\]
Taking 1 to the left hand side of the equation, we write,
\[\Rightarrow 1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}\]
Now, putting this value in our original equation, we can write it as,
\[\begin{align}
  & 1+\sec x+\cos x \\
 & \Rightarrow \left( 1+\cos x \right)+\sec x \\
 & \Rightarrow 2{{\cos }^{2}}\dfrac{x}{2}+\sec x \\
\end{align}\]
Thus, we can say that our given problem reduces to or simplifies to \[2{{\cos }^{2}}\dfrac{x}{2}+\sec x\] .

Note:
 We should always remember all the equations and general values of trigonometry or else solving such problems may be difficult. We can also solve this problem using another method, which includes replacing \[\sin =\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}\] and \[\cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}\] . But it is a pretty lengthy process, but yields the same answer.