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 - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 31


$$1+\tan x\tan2x=\sec2x$$ The equation is an identity since two sides are equal to each other as shown below.

Work Step by Step

$$1+\tan x\tan2x=\sec2x$$ We started dealing with the left side. $$X=1+\tan x\tan2x$$ - From Double-Angle Identity for tangent, we get $$\tan2x=\frac{2\tan x}{1-\tan^2x}$$ Thus, replace into $X$: $$X=1+\tan x\times\frac{2\tan x}{1-\tan^2x}$$ $$X=1+\frac{2\tan^2x}{1-\tan^2x}$$ Now, the right side has $\sec2x$, which is related to $\cos2x$, we probably should write $\tan x$ as $\frac{\sin x}{\cos x}$ here. $$X=1+\frac{2\times\frac{\sin^2x}{\cos^2x}}{1-\frac{\sin^2x}{\cos^2x}}$$ $$X=1+\frac{\frac{2\sin^2x}{\cos^2x}}{\frac{\cos^2x-\sin^2x}{\cos^2x}}$$ 2 $\cos^2x$ would eliminate each other. $$X=1+\frac{2\sin^2x}{\cos^2x-\sin^2x}$$ $$X=\frac{\cos^2x-\sin^2x+2\sin^2x}{\cos^2x-\sin^2x}$$ $$X=\frac{\cos^2x+\sin^2x}{\cos^2x-\sin^2x}$$ - From Pythagorean Identities: $\cos^2x+\sin^2x=1$ - From Double-Angle Identities for cosine: $\cos^2x-\sin^2x=\cos2x$ $$X=\frac{1}{\cos2x}$$ $$X=\sec2x\hspace{1cm}\text{(Reciprocal Identity)}$$ Therefore, $$1+\tan x\tan2x=\sec2x$$ We can conclude that the equation is an identity.
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