How do you simplify the identity: \[\dfrac{{\sec x - 1}}{{1 - \cos x}} = \sec x\] ?
Answer
575.7k+ views
Hint: In the given question, we have to verify that \[\dfrac{{\sec x - 1}}{{1 - \cos x}} = \sec x\] ,that is, we have to show that the part on the right side of the equal to sign is equal to the part on the left side of the equal to sign. In simple words, we have to prove that the right-hand side of the given equation is equal to the left-hand side. For that, we will take any one side and solve it using the trigonometric ratios or identities and make it equal to the other side.
Complete step-by-step solution:
We know that $\sec x = \dfrac{1}{{\cos x}}$ , so we get –
$ \Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \dfrac{{\dfrac{1}{{\cos x}} - 1}}{{1 - \cos x}}$
Now, we will take the LCM of the denominators in the numerator of the obtained term, and get –
$
\Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \dfrac{{\dfrac{{1 - \cos x}}{{\cos x}}}}{{1 - \cos x}} \\
\Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \dfrac{{1 - \cos x}}{{\cos x(1 - \cos x)}} \\
\Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \dfrac{1}{{\cos x}} \\
\Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \sec x \\
$
Now, the left-hand side has become equal to the right-hand side.
Hence, the simplified identity is $\dfrac{{\sec x - 1}}{{1 - \cos x}} = \sec x$ .
Note: The ratios of two sides of a right-angled triangle are known as trigonometric ratios. The ratio of perpendicular and the hypotenuse of the right-angled triangle is known as the sine function, the ratio of the base and the hypotenuse of the right-angled triangle is known as the cosine function and the ratio of the perpendicular and the base of the right-angled triangle is known as the tangent function.
The reciprocal of the sine, cosine and tangent function are known as cosecant, secant and cotangent functions respectively, thus all the trigonometric functions are interrelated with each other and one can be converted into another using this knowledge or the trigonometric identities like we have converted secant function into cosine function using $\sec x = \dfrac{1}{{\cos x}}$ .
Complete step-by-step solution:
We know that $\sec x = \dfrac{1}{{\cos x}}$ , so we get –
$ \Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \dfrac{{\dfrac{1}{{\cos x}} - 1}}{{1 - \cos x}}$
Now, we will take the LCM of the denominators in the numerator of the obtained term, and get –
$
\Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \dfrac{{\dfrac{{1 - \cos x}}{{\cos x}}}}{{1 - \cos x}} \\
\Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \dfrac{{1 - \cos x}}{{\cos x(1 - \cos x)}} \\
\Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \dfrac{1}{{\cos x}} \\
\Rightarrow \dfrac{{\sec x - 1}}{{1 - \cos x}} = \sec x \\
$
Now, the left-hand side has become equal to the right-hand side.
Hence, the simplified identity is $\dfrac{{\sec x - 1}}{{1 - \cos x}} = \sec x$ .
Note: The ratios of two sides of a right-angled triangle are known as trigonometric ratios. The ratio of perpendicular and the hypotenuse of the right-angled triangle is known as the sine function, the ratio of the base and the hypotenuse of the right-angled triangle is known as the cosine function and the ratio of the perpendicular and the base of the right-angled triangle is known as the tangent function.
The reciprocal of the sine, cosine and tangent function are known as cosecant, secant and cotangent functions respectively, thus all the trigonometric functions are interrelated with each other and one can be converted into another using this knowledge or the trigonometric identities like we have converted secant function into cosine function using $\sec x = \dfrac{1}{{\cos x}}$ .
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