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Question: Prove the following expression: \(\dfrac{\cos A}{1-\sin A}+\dfrac{\cos A}{1+\sin A}=2\sec A\)...

Prove the following expression: cos⁡A1−sin⁡A+cos⁡A1+sin⁡A=2sec⁡A\dfrac{\cos A}{1-\sin A}+\dfrac{\cos A}{1+\sin A}=2\sec A

Explanation

Solution

Hint: First we will start from LHS and take out the common term cosA from the expression. And then we will solve the expression further. After that we will use the formula (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} to simplify it further and then we will use the trigonometric formula 1−sin⁡2x=cos⁡2x1-{{\sin }^{2}}x={{\cos }^{2}}x and 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x, to prove LHS = RHS.

Complete step-by-step solution -
Let’s start our solution.
Let’s start solving from LHS.
=cos⁡A1−sin⁡A+cos⁡A1+sin⁡A =cos⁡A(11−sin⁡A+11+sin⁡A) \begin{aligned} & =\dfrac{\cos A}{1-\sin A}+\dfrac{\cos A}{1+\sin A} \\\ & =\cos A\left( \dfrac{1}{1-\sin A}+\dfrac{1}{1+\sin A} \right) \\\ \end{aligned}
Now after taking LCM we will add the two expression and then use the formula (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}
=cos⁡A(1+sin⁡A+1−sin⁡A(1+sin⁡A)(1−sin⁡A))=\cos A\left( \dfrac{1+\sin A+1-\sin A}{\left( 1+\sin A \right)\left( 1-\sin A \right)} \right)
Now using the formula (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} on (1+sin⁡A)(1−sin⁡A)\left( 1+\sin A \right)\left( 1-\sin A \right) we get,
=cos⁡A(21−sin⁡2A)=\cos A\left( \dfrac{2}{1-{{\sin }^{2}}A} \right)
Now using the formula 1−sin⁡2x=cos⁡2x1-{{\sin }^{2}}x={{\cos }^{2}}x we get,
=cos⁡A(2cos⁡2A) =2cos⁡A \begin{aligned} & =\cos A\left( \dfrac{2}{{{\cos }^{2}}A} \right) \\\ & =\dfrac{2}{\cos A} \\\ \end{aligned}
Now we know that 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x, hence using this in the above expression we get
=2sec⁡A=2\sec A
Hence, we have proved that LHS = RHS.
Hence, cos⁡A1−sin⁡A+cos⁡A1+sin⁡A=2sec⁡A\dfrac{\cos A}{1-\sin A}+\dfrac{\cos A}{1+\sin A}=2\sec A has been proved.

Note: To solve this we have used the algebraic formula (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} and some trigonometric formula 1−sin⁡2x=cos⁡2x1-{{\sin }^{2}}x={{\cos }^{2}}x and 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x, to prove LHS = RHS. Here we have started solving from LHS one can also solve this question by starting from RHS and then we have add and subtract something in both numerator and denominator to convert RHS into LHS. The process of converting RHS to LHS will be the total opposite that we have used to solve this question. But to convert RHS to LHS is quite hard so we must try to solve it from LHS and show it equals RHS.