Question

How do you prove $\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{\cos x}}{{1 - \sin x}}} \right) = 2\sec x$?

Answer 1

Hint: To prove this type of question you need to use basic trigonometric identities. To prove you have to derive R.H.S from L.H.S using trigonometric identities such as ${\sin ^2}x + {\cos ^2}x = 1$ and $\dfrac{1}{{\cos x}} = \sec x$. To prove this you have to first progress as you do in fraction addition and after result follows from it.

Complete step by step answer:
Let’s try to prove this trigonometric equation by using basic trigonometric identities.
We need to prove: $\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{\cos x}}{{1 - \sin x}}} \right) = 2\sec x$
Proof: We start with L.H.S of the equation and reduce it R.H.S of the equation. Since we have reduced one side of the equation to the other side, it means that both are equal and hence proved.
Starting with L.H.S $ = \left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{\cos x}}{{1 - \sin x}}} \right)$
 Now we will take the L.C.M (least common multiple) of the denominators and we get,
 L.C.M of $(1 + \sin x)$ and $(1 - \sin x)$ is equal to $(1 - {\sin ^2}x)$ and now perform fraction addition to reduce L.H.S to
 $ \Rightarrow L.H.S = \left( {\dfrac{{\cos x(1 - \sin x) + \cos x(1 + \sin x)}}{{1 - {{\sin }^2}x}}} \right)$ $eq(1)$
Now by simplifying numerator of $eq(1)$ we will get,
 $ \Rightarrow L.H.S = \left( {\dfrac{{\cos x - \cos x\sin x + \cos x + \cos x\sin x}}{{1 - {{\sin }^2}x}}} \right)$ $eq(2)$
After canceling $\cos x\sin x$ terms and adding $\cos x$ terms from $eq(2)$ we get
 $ \Rightarrow L.H.S = \left( {\dfrac{{2\cos x}}{{1 - {{\sin }^2}x}}} \right)$ $eq(3)$
And since we know that ${\sin ^2}x + {\cos ^2}x = 1$, we can also write this identity as
$ \Rightarrow (1 - {\sin ^2}x) = {\cos ^2}x$.
Replacing $(1 - {\sin ^2}x)$in $eq(3)$we get,
 $ \Rightarrow L.H.S = \left( {\dfrac{{2\cos x}}{{{{\cos }^2}x}}} \right)$ $eq(4)$
Since we have $\cos x$ both in denominator and numerator of $eq(4)$ we cancel one power of $\cos x$ from both numerator and denominator. So our equation become
 $ \Rightarrow L.H.S = \left( {\dfrac{2}{{\cos x}}} \right)$ $eq(5)$
And since we know that $\dfrac{1}{{\cos x}} = \sec x$, by using trigonometric identity in $eq(5)$ we get,
$ \Rightarrow L.H.S = 2\sec x = R.H.S$
Since we have got R.H.S from the L.H.S of the given equation.
Hence proved, $\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{\cos x}}{{1 - \sin x}}} \right) = 2\sec x$

Note:We can also with R.H.S of the equation to derive L.H.S of the equation, but we usually start with the L.H.S of the equation. To prove this type of question based on trigonometric identities you must have to learn basic identities.