Solveeit Logo
Login

Question

Question: How do you verify: \(\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x\)?...

How do you verify: cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=2sec⁡x\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x?

Explanation

Solution

We have sum of two terms in the left-hand side of cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=2sec⁡x\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x. We take the LCM of the denominators. Then we add them in the denominators. Then we use the identity of (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab to simplify the numerator and also use sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1.

Complete step-by-step solution:
We have the sum of two terms in cos⁡x1+sin⁡x+1+sin⁡xcos⁡x\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}. We take the LCM of the denominators as the multiplications of those terms.
The equation becomes cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=cos⁡2x+(1+sin⁡x)2cos⁡x(1+sin⁡x)\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{{{\cos }^{2}}x+{{\left( 1+\sin x \right)}^{2}}}{\cos x\left( 1+\sin x \right)}.
We break the square as (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab. So, (1+sin⁡x)2=1+sin⁡2x+2sin⁡x{{\left( 1+\sin x \right)}^{2}}=1+{{\sin }^{2}}x+2\sin x.
The equation becomes cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=cos⁡2x+(1+sin⁡x)2cos⁡x(1+sin⁡x)=cos⁡2x+1+sin⁡2x+2sin⁡xcos⁡x(1+sin⁡x)\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{{{\cos }^{2}}x+{{\left( 1+\sin x \right)}^{2}}}{\cos x\left( 1+\sin x \right)}=\dfrac{{{\cos }^{2}}x+1+{{\sin }^{2}}x+2\sin x}{\cos x\left( 1+\sin x \right)}.
We know that sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. We get cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=2+2sin⁡xcos⁡x(1+sin⁡x)\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{2+2\sin x}{\cos x\left( 1+\sin x \right)}.
We take 2 common from the numerator to get cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=2(1+sin⁡x)cos⁡x(1+sin⁡x)\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{2\left( 1+\sin x \right)}{\cos x\left( 1+\sin x \right)}.
We omit the common from both numerator and denominator and get
cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=2(1+sin⁡x)cos⁡x(1+sin⁡x)=2cos⁡x\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=\dfrac{2\left( 1+\sin x \right)}{\cos x\left( 1+\sin x \right)}=\dfrac{2}{\cos x}.
We use the inverse formula to get 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x. So, cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=2sec⁡x\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x.
Thus verified cos⁡x1+sin⁡x+1+sin⁡xcos⁡x=2sec⁡x\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x.

Note: It is important to remember that the condition to eliminate the (1+sin⁡x)\left( 1+\sin x \right) from both denominator and numerator is (1+sin⁡x)≠0\left( 1+\sin x \right)\ne 0. No domain is given for the variable xx. The value of sin⁡x≠−1\sin x\ne -1 is essential. The simplified condition will be x≠(4n−1)π2,n∈Zx\ne \dfrac{\left( 4n-1 \right)\pi }{2},n\in \mathbb{Z}.