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How do you find the exact value of $\cot \left( \dfrac{5\pi }{4} \right)$ ?

Last updated date: 01st Mar 2024
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IVSAT 2024
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Hint: To solve the above question use the property of the trigonometric functions that they periodic functions and have a period of $2\pi $. Write the given angle as the sum of $2\pi $ and some other angle and proceed further.

Complete step by step solution:
Let us first understand the procedure to find the exact value of any trigonometric ratios.We know that the basic trigonometric ratios are sine and cosine and all the other trigonometric ratios depend on these two ratios. We also know that functions of sine and cosine are periodic functions (meaning their values repeat after equal intervals). The sine and cosine functions have a period of $2\pi $. And similarly, all the other trigonometric functions have a period of $2\pi $. This means that any trigonometric function will repeat itself after every equal interval of $2\pi $ units. If we plot the given trigonometric function, we can see this property more clearly.

Suppose, we have a trigonometric function, say sin(x) and let the value of the function sin(x), at $x={{x}_{0}}$ has some value, say y. Then the function will have the same value (i.e. y) at $x={{x}_{0}}+2\pi $, since the function has a period of $2\pi $.This means that $\sin (2\pi +{{x}_{0}})=\sin ({{x}_{0}})$.This relation is applicable for all the trigonometric ratios.In the question, the given function is cot(x).The value $\dfrac{5\pi }{4}$ can be written as $\dfrac{5\pi }{4}=2\pi -\dfrac{3\pi }{4}$.
Then, this means that $\cot \left( \dfrac{5\pi }{4} \right)=\cot \left( 2\pi -\dfrac{3\pi }{4} \right)$.
And we know that $\cot \left( 2\pi -\dfrac{3\pi }{4} \right)=\cot \left( -\dfrac{3\pi }{4} \right)$.
We also know that $\cot \left( -\dfrac{3\pi }{4} \right)=1$

Therefore, the exact value of $\cot \left( \dfrac{5\pi }{4} \right)$ is 1.

Note: To solve the given problem we can also use the relation $\cot (\pi +\theta )=\cot (\theta )$. However, this relation is only valid for the trigonometric functions – tan(x) and cot(x). For sine and cosine functions,
$\sin (\pi +\theta )=-\sin (\theta )$
$\cos (\pi +\theta )=-\cos (\theta )$
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