Question

If we have a trigonometric equation as$\left( 1+\cos A \right)\left( 1-\cos A \right)=\dfrac{3}{4}$, find the value of $\sec A$.

Answer 1

Hint: In this given question, we may first use the formula of$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ to convert $\left( 1+\cos A \right)\left( 1-\cos A \right)$ into the form of ${{1}^{2}}-{{\cos }^{2}}A$ then get the value of θ by using the corollary of the trigonometric formula for the sum of the squares of sine and cosine of an angle equalizes to 1 that gives us the value of the square of the sine of an angle is equal to the difference of 1 and the square of cosine of the angle. That is,
$\begin{aligned} & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & \Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \\\ \end{aligned}$

Complete step-by-step solution -
In this given question, we are asked to find out the value of $\sec A$if$\left( 1+\cos A \right)\left( 1-\cos A \right)=\dfrac{3}{4}$.
Here, we are going to use the formula of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}..............(1.1)$.
Then we will get the value of θ by using the corollary of the trigonometric formula for the sum of the squares of sine and cosine of an angle equalizes to 1 that gives us the value of the square of the sine of an angle is equal to the difference of 1 and the square of cosine of the angle. That is,
$\begin{aligned} & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & \Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta .................(1.2) \\\ \end{aligned}$
The process is as follows:
By using equation 1.1, we get,
$\begin{aligned} & \left( 1+\cos A \right)\left( 1-\cos A \right)=\dfrac{3}{4} \\\ & \Rightarrow {{1}^{2}}-{{\cos }^{2}}A=\dfrac{3}{4} \\\ & \Rightarrow {{\cos }^{2}}A=1-\dfrac{3}{4} \\\ & \Rightarrow {{\cos }^{2}}A=\dfrac{1}{4} \\\ & \Rightarrow \cos A=\pm \dfrac{1}{2}...............(1.3) \\\ \end{aligned}$
Now, as we know the value of $\cos {{60}^{\circ }}=\dfrac{1}{2}\text{ and }\cos \left( {{120}^{\circ }} \right)=\dfrac{-1}{2}...............(1.4)$.
So, we can say that angle A is equal to ${{60}^{\circ }}$ or ${{120}^{\circ }}$ from equation 1.3 and 1.4.
So, if $A={{60}^{\circ }}$ , $\sec A=\sec {{60}^{\circ }}=2$ and
If $A={{120}^{\circ }}$ , $\sec A=\sec {{120}^{\circ }}=-2$
Hence, we have got our answer as 2 or -2 as the value of $\sec A$.

Note: We should note that in equation (1.4), we equated $\theta$ to ${{60}^{\circ }}=\dfrac{\pi }{3}$ or ${{120}^{\circ }}=\dfrac{2\pi }{3}$ because the cosine of both these angles was the same. However, in the general case, as the cosine function is even function and has a periodicity of $2\pi$, the general value of $\theta$ should be $2n\pi \pm \dfrac{\pi }{3}$ or $2n\pi \pm \dfrac{2\pi }{3}$ . However, increasing an angle by $2\pi$ returns the vectors to their original position and as we are just asked about the angle and not its orientation, positive or negative angles represent the same thing. Thus, equating the angle $\theta$ to ${{60}^{\circ }}$ or ${{120}^{\circ }}$ in equation (1.4) is justified.