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Chapter 5: Problem 45

Verify each identity. $$(\sec x-\tan x)^{2}=\frac{1-\sin x}{1+\sin x}$$

Short Answer

Expert verified
The left side of the equation simplifies to match the right side, so the identity, \((\sec x-\tan x)^{2}=\frac{1-\sin x}{1+\sin x}\) is verified.

Step by step solution

01

Convert functions to Sine and Cosine

The first step is to convert the secant and tangent functions on the left side of the equation into functions of sine and cosine. Secant of x is equal to \(1/cosx\) and tangent of x is equal to \(sinx/cosx\). So, the left side of the equation becomes \((1/cosx - sinx/cosx)^2\) which simplifies to \((1-sinx)^2/(cosx)^2\).
02

Simplify the denominator by using Pythagorean Identity

The Pythagorean identity is \(sin^2(x) + cos^2(x) = 1\). Solve this for \(cos^2(x)\) to get \(cos^2(x)=1-sin^2(x)\). Substituting this into the equation, gives us \((1-sin^2(x))/(1-sin^2(x))\) for the left side of the equation.
03

Final Step: Simplify the Equation

Simplify the left side of the equation get \(1-sin^2(x)\). Then, factor out \((1-sin(x))(1+sin(x))\). Now the left side matches with the right side. Therefore, the given identity is verified.

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Most popular questions from this chapter

Verify the identity: $$\frac{\sin (x-y)}{\cos x \cos y}+\frac{\sin (y-z)}{\cos y \cos z}+\frac{\sin (z-x)}{\cos z \cos x}=0$$

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