Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Summary Exercises on Verifying Trigonometric Identities - Page 245: 17


$$\frac{\tan^2t+1}{\tan t\csc^2t}=\tan t$$ The equation is an identity. The proof is explained in detail in the work step by step section below.

Work Step by Step

$$\frac{\tan^2t+1}{\tan t\csc^2t}=\tan t$$ The left side would be examined first. $$X=\frac{\tan^2t+1}{\tan t\csc^2t}$$ Here we would consider the numerator and denominator separately. *Numerator: According to Pythagorean Identity: $$\tan^2t+1=\sec^2t$$ However, as $\sec t=\frac{1}{\cos t}$, we have $$\tan^2t+1=\sec^2t=\frac{1}{\cos^2t}$$ *Denominator: Apply the following identities: $$\tan t=\frac{\sin t}{\cos t}\hspace{2cm}\csc t=\frac{1}{\sin t}$$ we have $$\tan t\csc^2t=\frac{\sin t}{\cos t}\times\frac{1}{\sin^2t}$$ $$\tan t\csc^2t=\frac{1}{\sin t\cos t}$$ Combine the numerator and denominator back to $X$: $$X=\frac{\frac{1}{\cos^2t}}{\frac{1}{\sin t\cos t}}$$ $$X=\frac{\sin t\cos t}{\cos^2t}$$ $$X=\frac{\sin t}{\cos t}$$ - As we know, from Quotient Identity, $\tan t=\frac{\sin t}{\cos t}$, so $$X=\tan t$$ In conclusion, $$\frac{\tan^2t+1}{\tan t\csc^2t}=\tan t$$ The equation is an identity.
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