Question

Prove that: $\sec x\sqrt{1-{{\sin }^{2}}x}=1$

Answer 1

To prove the given trigonometric equation we will use relation between trigonometric functions. Firstly we will find the value of the term inside the bracket by using the square relation between sine and cosine function. Then we will change the term outside the bracket by using reciprocal relation between cosine and secant and so that we can make the left side equal to the right side and prove our equation.

Complete step-by-step solution:
We have to prove the below trigonometric equation:
$\sec x\sqrt{1-{{\sin }^{2}}x}=1$……$\left( 1 \right)$
So we will solve the left hand side and make it equal to the value at right hand side
Now, we know the square relation between sine and cosine function is as below:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
On rearranging the terms we get,
${{\cos }^{2}}x=1-{{\sin }^{2}}x$
On substituting above value in left side of equation (1) we get,
$\begin{aligned} & \Rightarrow \sec x\sqrt{{{\cos }^{2}}x} \\\ & \Rightarrow \sec x{{\left( \cos x \right)}^{\dfrac{1}{2}\times 2}} \\\ \end{aligned}$
$\therefore \sec x\left( \cos x \right)$……$\left( 2 \right)$
Next we know the reciprocal relation between the cosine and secant is:
$\sec x=\dfrac{1}{\cos x}$
Put above value in equation (2) and simplify:
$\begin{aligned} & \Rightarrow \dfrac{1}{\cos x}\times \cos x \\\ & \Rightarrow 1 \\\ \end{aligned}$
So we got the value as 1 which I equal to the right hand side
Hence it is proved that $\sec x\sqrt{1-{{\sin }^{2}}x}=1$

Note: Trigonometric functions are those real functions which are related to a right-angled triangle. These functions relate an angle of right-angle triangle to the ratio of its length of two sides. They are used in many fields of mathematics such as in geometry for navigation and geodesy. They are the simplest periodic functions that are used to study the periodic phenomena. All these functions are related to each other in one or the other way and these relations can be of square or reciprocal and many others.