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Chapter 5: Problem 18

Verify the identity. $$\frac{\cot ^{3} t}{\csc t}=\cos t\left(\csc ^{2} t-1\right)$$

Short Answer

Expert verified
The given trigonometric identity has been verified correctly, thus the identity \(\frac{\cot ^{3} t}{\csc t}=\cos t\left(\csc ^{2} t-1\right)\) holds true for all legal values of the variable \(t\).

Step by step solution

01

Express the LHS in Terms of One Trigonometric Function

Express cotangent and cosecant in terms of sine and cosine.Remember that \(\cot t = \frac{\cos t}{\sin t}\) and \(\csc t = \frac{1}{\sin t}\), thus, the left hand side \(LHS = \frac{\cot ^{3} t}{\csc t} = \frac{\left(\frac{\cos ^{3} t}{\sin ^{3} t}\right)}{\left(\frac{1}{\sin t}\right)}\).
02

Simplify the LHS

Simplify the expression obtained in the previous step. This gives us \(LHS = \frac{\cos ^{3} t}{\sin ^{2} t}\).
03

Express RHS in terms of sin and cos

Express the right hand side (RHS) in terms of sine and cosine functions. We have \(\csc ^{2} t = \frac{1}{\sin ^{2} t}\), thus the equation becomes \(RHS = \cos t\left(\csc ^{2} t-1\right) = \cos t \left(\frac{1}{\sin ^{2} t} - 1\right)\).
04

Further Simplifying RHS

Multiply through the brackets in the RHS. This gives us \(RHS =\frac{\cos t}{\sin ^{2} t} - \cos t\).
05

Comparing both sides of the equation

On comparing, the LHS and RHS are not yet the same. Notice that the second part of the RHS (\(- \cos t\)) is equal to \(- \frac{\cos t \sin ^{2} t}{\sin ^{2} t}\). This allows us to combine the terms in the RHS under the same denominator to finally get \(RHS = \frac{\cos t - \cos t \sin ^{2} t}{\sin ^{2} t} = \frac{\cos ^{3} t}{\sin ^{2} t}\), which is exactly the same as the left hand side. Thus, the given identity has been verified correctly.

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