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Chapter 19: Problem 1803

If \(2 \sec ^{2} \alpha-\sec ^{4} \alpha-2 \operatorname{cosec}^{2} \alpha+\operatorname{cosec}^{4} \alpha=(15 / 4)\) then \(\tan ^{2} \alpha=\) (a) \((1 / \sqrt{2})\) (b) \((1 / 2)\) (c) \([1 /(2 \sqrt{2})]\) (d) \((1 / 4)\)

Short Answer

Expert verified
The correct answer is: \( \tan^2\alpha = \frac{1}{2} \)

Step by step solution

01

Recall that the secant function is the reciprocal of the cosine function, and the cosecant function is the reciprocal of the sine function. The given equation is: \(2\sec^2\alpha - \sec^4\alpha - 2\cosec^2\alpha + \cosec^4\alpha = \frac{15}{4}\) We can rewrite this equation using the identities \( \sec\alpha = \frac{1}{\cos\alpha} \) and \( \cosec\alpha = \frac{1}{\sin\alpha} \): \(2\left(\frac{1}{\cos^2\alpha}\right) - \left(\frac{1}{\cos^4\alpha}\right) - 2\left(\frac{1}{\sin^2\alpha}\right) + \left(\frac{1}{\sin^4\alpha}\right) = \frac{15}{4}\) #Step 2: Simplify the equation#

Simplify the left side of the equation by combining terms involving sine and cosine functions: \(\frac{2 - \cos^2\alpha}{\cos^4\alpha} - \frac{2 - \sin^2\alpha}{\sin^4\alpha} = \frac{15}{4}\) Now, recall that \( \sin^2\alpha + \cos^2\alpha = 1 \). We can rewrite the left side of the equation using this identity: \(\frac{1 + \sin^2\alpha - \sin^2\alpha\cos^2\alpha}{\cos^4\alpha\sin^4\alpha} = \frac{15}{4}\) #Step 3: Use the relationship between sine, cosine, and tangent functions to transform the equation#
02

Recall that \( \tan\alpha = \frac{\sin\alpha}{\cos\alpha} \) or \( \tan^2\alpha = \frac{\sin^2\alpha}{\cos^2\alpha} \). Substitute this into the equation: \(\frac{1 + \tan^2\alpha - \tan^2\alpha\cos^2\alpha\sin^2\alpha}{\cos^4\alpha\sin^4\alpha} = \frac{15}{4}\) Now, multiply both sides of the equation by \( \cos^4\alpha\sin^4\alpha \): \(1 + \tan^2\alpha - \tan^2\alpha\cos^2\alpha\sin^2\alpha = \frac{15}{4}\cos^4\alpha\sin^4\alpha\) Then subtract 1 from both sides: \(\tan^2\alpha - \tan^2\alpha\cos^2\alpha\sin^2\alpha = \frac{15}{4}\cos^4\alpha\sin^4\alpha - 1\) #Step 4: Test the given answer choices#

Now, we can test the answer choices in the equation: (a) \( \tan^2\alpha = \frac{1}{\sqrt{2}} \) (b) \( \tan^2\alpha = \frac{1}{2} \) (c) \( \tan^2\alpha = \frac{1}{2\sqrt{2}} \) (d) \( \tan^2\alpha = \frac{1}{4} \) After testing each of the answer choices, we find that the correct option is: (b) \( \tan^2\alpha = \frac{1}{2} \)

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Most popular questions from this chapter

The angle of depression of the top and bottom of a tower observed from top of a lighthouse of 300 meter height are \(30^{\circ}\) and \(60^{\circ}\) respectively then the height of the tower is (a) 300 meter (b) \(100 \mathrm{~m}\) (c) \(200 \mathrm{~m}\) (d) \(50 \mathrm{~m}\)

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