Question

Prove that: $\dfrac{\cos \theta }{1+\sin \theta }+\dfrac{1+\sin \theta }{\cos \theta }=2\sec \theta $ .

Answer 1

Hint: The terms in the left-hand side of our expression contain ‘sine’ and ‘cosine’ terms whereas the term in the right-hand side of our expression is a ‘secant’. So, we can convert the secant term into its ‘cosine’ form by taking its reciprocal and then work simultaneously on the left-hand expression. This will be the base of our proof and we shall proceed in this manner only.

Complete step-by-step solution:
We have been given the expression to proof as: $\dfrac{\cos \theta }{1+\sin \theta }+\dfrac{1+\sin \theta }{\cos \theta }=2\sec \theta $.
First of all, working on the right-hand side of our expression, we get:
$\begin{align}
  & \Rightarrow R.H.S.=2\sec \theta \\
 & \therefore R.H.S.=\dfrac{2}{\cos \theta } \\
\end{align}$
Let us say the above equation is equation number (1). So, we have:
$\Rightarrow R.H.S.=\dfrac{2}{\cos \theta }$
Now, working on the left-hand side of our expression, we get:
$\begin{align}
  & \Rightarrow L.H.S.=\dfrac{\cos \theta }{1+\sin \theta }+\dfrac{1+\sin \theta }{\cos \theta } \\
 & \Rightarrow L.H.S.=\dfrac{\cos \theta \left( \cos \theta \right)+\left( 1+\sin \theta \right)\left( 1+\sin \theta \right)}{\left( 1+\sin \theta \right)\times \left( \cos \theta \right)} \\
 & \Rightarrow L.H.S.=\dfrac{{{\cos }^{2}}\theta +1+2\sin \theta +{{\sin }^{2}}\theta }{\left( 1+\sin \theta \right)\times \left( \cos \theta \right)} \\
\end{align}$
Now, the trigonometric identity relating “sine” and “cosine” terms can be written as follows:
$\Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A=1$
Using this identity, our equation can be further simplified into:
$\begin{align}
  & \Rightarrow L.H.S.=\dfrac{1+1+2\sin \theta }{\left( 1+\sin \theta \right)\times \left( \cos \theta \right)} \\
 & \Rightarrow L.H.S.=\dfrac{2+2\sin \theta }{\left( 1+\sin \theta \right)\times \left( \cos \theta \right)} \\
 & \Rightarrow L.H.S.=\dfrac{2\left( 1+\sin \theta \right)}{\left( 1+\sin \theta \right)\times \left( \cos \theta \right)} \\
\end{align}$
Now, cancelling out common terms from numerator and denominator, we get:
$\Rightarrow L.H.S.=\dfrac{2}{\cos \theta }$
Let us say this is equation number (2). So, we have:
$\Rightarrow L.H.S.=\dfrac{2}{\cos \theta }$ .......... (2)
From equation number (1) and (2), we can say that:
$\Rightarrow L.H.S.=R.H.S.$
Therefore, $\dfrac{\cos \theta }{1+\sin \theta }+\dfrac{1+\sin \theta }{\cos \theta }=2\sec \theta $ is true.
Hence, the given expression has been proved.

Note: Whenever going for a proof in a problem, we should try and manipulate both sides of our expression. This is done to simplify both the sides, so that our proof becomes comparatively easier. Also, in trigonometry, the identities are the backbone of every problem. So, one should remember all the trigonometric identities thoroughly, so that there is no need to derive them in an exam, as this will help save a lot of time.