Problem 966 Prove the identity \(\sin ^{2} \... [FREE SOLUTION]

Chapter 29: Problem 966

Prove the identity \(\sin ^{2} \theta+\tan ^{2} \theta=\sec ^{2} \theta-\cos ^{2} \theta\).

Short Answer

Expert verified

We can prove the given identity \(\sin^2 \theta + \tan^2 \theta = \sec^2 \theta - \cos^2 \theta\) by converting all the trigonometric functions on the LHS into functions involving sine and cosine, and the RHS into functions involving secant and cosine. After simplifying, we find that the LHS equals to \(\frac{\sin^2 \theta \cdot \cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}\) while the RHS equals to \(\frac{1 - \cos^4 \theta}{\cos^2 \theta}\). By the Pythagorean identity \(\sin^2 \theta = 1 - \cos^2 \theta\), the RHS expression becomes equal to the LHS expression, proving the given identity.

Step by step solution

Convert trigonometric functions on LHS to sine and cosine

LHS: \(\sin^2 \theta + \tan^2 \theta\) We know that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Therefore, LHS becomes: \(\sin^2 \theta + \left(\frac{\sin \theta}{\cos \theta}\right)^2\)

Simplify LHS

Now, we simplify the expression: LHS: \(\sin^2 \theta + \frac{\sin^2 \theta}{\cos^2 \theta}\) To simplify further, we need a common denominator for the two terms. The common denominator is \(\cos^2 \theta\): LHS: \(\frac{\sin^2 \theta \cdot \cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}\)

Convert trigonometric functions on RHS to secant and cosine

RHS: \(\sec^2 \theta - \cos^2 \theta\) We know that \(\sec \theta = \frac{1}{\cos \theta}\). Therefore, RHS becomes: \(\left(\frac{1}{\cos \theta}\right)^2 - \cos^2 \theta\)

Simplify RHS

Now, we simplify the expression: RHS: \(\frac{1}{\cos^2 \theta} - \cos^2 \theta\) Again, we need a common denominator for the two terms. The common denominator is \(\cos^2 \theta\): RHS: \(\frac{1 - \cos^4 \theta}{\cos^2 \theta}\)

Compare LHS and RHS

Let's compare the LHS and RHS expressions: \(LHS = \frac{\sin^2 \theta \cdot \cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}\) \(RHS = \frac{1 - \cos^4 \theta}{\cos^2 \theta}\) Notice that \(1 - \cos^2 \theta = \sin^2 \theta\), making the RHS expression equivalent to: \[\frac{\sin^2 \theta \cdot \cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}\] Thus, LHS = RHS, proving the identity: \(\sin^2 \theta + \tan^2 \theta = \sec^2 \theta - \cos^2 \theta\)

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91影视

Prove the identity \(\sin ^{2} \theta+\tan ^{2} \theta=\sec ^{2} \theta-\cos ^{2} \theta\).

Short Answer

Step by step solution

Convert trigonometric functions on LHS to sine and cosine

Simplify LHS

Convert trigonometric functions on RHS to secant and cosine

Simplify RHS

Compare LHS and RHS

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