Question

How do you prove $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x$?

Answer 1

In this question, we are given a trigonometric equation and we need to prove it. For this, we will pick the left side of the equation and solve it to make it equal to the right side of the equation. We will first take LCM from the denominator in the left side and then solve the left side using various algebraic and trigonometric identities. The identities that we will use are,
(I) ${{\left( a+b \right)}^{2}}$ is equal to ${{a}^{2}}+{{b}^{2}}+2ab$.
(II) ${{\cos }^{2}}x+{{\sin }^{2}}x$ is equal to 1.
(III) Reciprocal of cosx is secx.

Complete step by step answer:
Here we are given the trigonometric equation as $\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x$.
We need to prove it.
For this, let us take the left side of the equation and try to prove it as equal to the right side.
$\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}$.
Let us take the least common multiple in the denominator as $\cos x\left( 1+\sin x \right)$. After suitable multiplication the single fraction becomes $\dfrac{{{\cos }^{2}}x+{{\left( 1+\sin x \right)}^{2}}}{\cos x\left( 1+\sin x \right)}$.
As we know from the algebraic identity that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So let us apply it on ${{\left( 1+\sin x \right)}^{2}}$. We get the left side of the equation as $\dfrac{{{\cos }^{2}}x+\left( 1+{{\sin }^{2}}x+2\sin x \right)}{\cos x\left( 1+\sin x \right)}$.
Opening up the bracket and rearranging the terms we get $\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x+1+2\sin x}{\cos x\left( 1+\sin x \right)}$.
Now as we know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. So let us use it in the numerator of the expression we get $\dfrac{1+1+2\sin x}{\cos x\left( 1+\sin x \right)}$.
Adding the constant we get $\dfrac{2+2\sin x}{\cos x\left( 1+\sin x \right)}$.
Let us take 2 common from the numerator: we get $\dfrac{2\left( 1+\sin x \right)}{\cos x\left( 1+\sin x \right)}$.
As we can see that, both the numerator and the denominator have a common factor $\left( 1+\sin x \right)$. So let us remove it. We are left with the expression as $\dfrac{2}{\cos x}$.
Now let's use the trigonometric identity according to which cosx is the reciprocal of secx. So we can say that $\dfrac{1}{\cos x}=\sec x$. Thus the expression becomes 2secx.
As we can see this is equal to the right side.
Hence we have proved the given trigonometric equation.

Note:
Students should keep in mind all the trigonometric as well as algebraic identities used to solve this sum. Note that, to apply ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ angle of both cosine and sine function must be the same. Make sure that factor is identical while calculating it from the numerator and the denominator.