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Chapter 5: Problem 41

Verify the identity. $$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}=\frac{1+\sin \theta}{|\cos \theta|}$$

Short Answer

Expert verified
Yes, the given identity \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta }} = \frac{1+\sin \theta}{| \cos \theta |} \) is verified to be correct.

Step by step solution

01

Start with one side of the equation

Start with the left side (LS) of the identity which is: \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}}\)
02

Rationalize the denominator

Rationalize the denominator by multiplying numerator and denominator inside the square root by the conjugate of the denominator. The conjugate of \(1-\sin \theta\) is \(1+\sin \theta\), giving \(\sqrt{\frac{(1+\sin \theta)^2}{1-\sin^2 \theta}}\)
03

Simplify and use trigonometric identity

Simplify to get \(\sqrt{\frac{1+2\sin \theta +\sin^2 \theta}{1-\sin^2 \theta}}.\) Then, use Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), which implies that \(1-\sin^2 \theta = \cos^2 \theta\), giving: \( \sqrt{\frac{1+2\sin \theta +\sin^2 \theta}{\cos^2 \theta}}\)
04

Split the square root

Split the fraction and then take the square root of each part, resulting in \(\frac{ \sqrt{1+2\sin \theta+\sin^2 \theta}}{|\cos \theta|}\)
05

Remove the absolute value

The denominator is \( |\cos \theta| \). Removing the absolute value gives two options, \( \cos \theta \) and \( -\cos \theta \). Since \( \cos \theta \) in denominator wouldn't make the LS equal to the RHS, the correct option must be \( -\cos \theta \). This results in LS equals to \( -\frac{ \sqrt{1+2\sin \theta+\sin^2 \theta}}{\cos \theta}\)
06

Simplify numerator

Simplify the numerator \( \sqrt{1+2\sin \theta+\sin^2 \theta}\) by writing the terms inside the square root as a perfect square. This would equate to \( -\frac{(1+\sin \theta) }{\cos \theta}\), making the left side (LS) same as the right side (RS)

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