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Question: Prove that \(\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\)?...

Prove that cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A?

Explanation

Solution

Hint : We have sum of two terms in the left-hand side of cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A. 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 answer :
We have the sum of two terms in cos⁡A1+sin⁡A+1+sin⁡Acos⁡A\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}. We take the LCM of the denominators as the multiplications of those terms.
The equation becomes
cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=cos⁡2A+(1+sin⁡A)2cos⁡A(1+sin⁡A)\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+{{\left( 1+\sin A \right)}^{2}}}{\cos A\left( 1+\sin A \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⁡A)2=1+sin⁡2A+2sin⁡A{{\left( 1+\sin A \right)}^{2}}=1+{{\sin }^{2}}A+2\sin A.
The equation becomes
cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=cos⁡2A+(1+sin⁡A)2cos⁡A(1+sin⁡A)=cos⁡2A+1+sin⁡2A+2sin⁡Acos⁡A(1+sin⁡A)\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+{{\left( 1+\sin A \right)}^{2}}}{\cos A\left( 1+\sin A \right)}=\dfrac{{{\cos }^{2}}A+1+{{\sin }^{2}}A+2\sin A}{\cos A\left( 1+\sin A \right)}.
We know that
sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. We get cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2+2sin⁡Acos⁡A(1+sin⁡A)\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2+2\sin A}{\cos A\left( 1+\sin A \right)}.
We take 2 common from the numerator to get
cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2(1+sin⁡A)cos⁡A(1+sin⁡A)\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2\left( 1+\sin A \right)}{\cos A\left( 1+\sin A \right)}.
We omit the common from both numerator and denominator and get
cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2(1+sin⁡A)cos⁡A(1+sin⁡A)=2cos⁡A\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2\left( 1+\sin A \right)}{\cos A\left( 1+\sin A \right)}=\dfrac{2}{\cos A}.
We use the inverse formula to get
1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x. So, cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2cos⁡A=2sec⁡A\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2}{\cos A}=2\sec A.
Thus verified cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A.

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