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Chapter 19: Problem 1865

\(\cos (x-y)=a, \cos (x+y) \Rightarrow \cot x \cot y=\) (a) \([(a-1) /(a+1)]\) (b) \([(a+1) /(a-1)]\) (c) \(a-1\) (d) \(a+1\)

Short Answer

Expert verified
The short answer is: \(\cot{x} \cot{y} = \frac{(a+1)}{(a-1)}\).

Step by step solution

01

Write down useful trigonometric identities

In this problem, we will need to use two product-to-sum trigonometric identities: 1. \(\cos{A} \cos{B} = \frac{1}{2}[\cos(A-B) + \cos(A+B)]\) 2. \(\sin{A} \sin{B} = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\) We will also need the relationships of cotangent with cosine and sine: 3. \(\cot{A} = \frac{\cos{A}}{\sin{A}}\)
02

Express cotangent in terms of cosine and sine

Write down the expression for \(\cot{x} \cot{y}\) using relationship 3: \[\cot{x} \cot{y} = \frac{\cos{x}\cos{y}}{\sin{x}\sin{y}}\]
03

Use product-to-sum formulas

Substitute the product expressions for the cosines and sines in the numerator and denominator of the cotangent expression using the product-to-sum identities 1 and 2: \[\cot{x} \cot{y} = \frac{\frac{1}{2}[\cos(x-y) + \cos(x+y)]}{\frac{1}{2}[\cos(x-y) - \cos(x+y)]}\]
04

Cancel out common terms and simplify

Cancel out the common term \(\frac{1}{2}\) in the numerator and denominator, and substitute the given value of \(\cos(x-y) = a\) in the expression: \[\cot{x} \cot{y} = \frac{a + \cos(x+y)}{a - \cos(x+y)}\]
05

Compare with the given options

Compare the simplified expression with the given options. The expression obtained in Step 4 matches option (b): \(\cot{x} \cot{y} = [(a+1)/(a-1)]\) Therefore, the correct option is (b).

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Most popular questions from this chapter

There is a bridge of the length \(h\) on a valley. The angle of depression of a temple lying in a valley from two ends of a bridge are \(\alpha\) and \(\beta\), then the height of the bridge from top of the temple \(=\) (a) \([(h \tan \alpha \tan \beta) /(\tan \alpha-\tan \beta)]\) (b) \([(h \tan \alpha \tan \beta) /(\tan \alpha+\tan \beta)]\) (c) \([(\tan \alpha \tan \beta) /\\{h(\tan \alpha-\tan \beta)\\}]\) (d) \([\\{h(\tan \alpha+\tan \beta)\\} /(\tan \alpha \tan \beta)]\)

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