Calculating the Value of cot^{-1}7 cot^{-1}8 cot^{-1}18 Using Inverse Trigonometric Identities

Introduction

In this article, we will explore how to solve the complex expression involving the product of three inverse cotangent functions: cot^{-1}7 , cot^{-1}8 , cot^{-1}18. We will utilize inverse trigonometric identities to simplify and find the value of this expression.

Step-by-Step Solution

To solve the expression cot^{-1}7 , cot^{-1}8 , cot^{-1}18, we can use the identity for the sum of inverse cotangents:

cot^{-1}x , cot^{-1}y cot^{-1}left(dfrac{xy - 1}{x y}right)

Step 1: Combine cot^{-1}7 and cot^{-1}8

First, we combine cot^{-1}7 and cot^{-1}8 using the identity:

cot^{-1}7 , cot^{-1}8 cot^{-1}left(dfrac{7 cdot 8 - 1}{7 8}right)

Perform the calculations:

Num: 7 cdot 8 - 1 56 - 1 55

Den: 7 8 15

Therefore:

cot^{-1}7 , cot^{-1}8 cot^{-1}left(dfrac{55}{15}right) cot^{-1}left(dfrac{11}{3}right)

Step 2: Combine cot^{-1}left(dfrac{11}{3}right) and cot^{-1}18

Next, we combine cot^{-1}left(dfrac{11}{3}right) and cot^{-1}18 using the same identity:

cot^{-1}left(dfrac{11}{3}right) , cot^{-1}18 cot^{-1}left(dfrac{left(dfrac{11}{3}right) cdot 18 - 1}{left(dfrac{11}{3}right) 18}right)

Calculate the numerator:

Num: left(dfrac{11}{3}right) cdot 18 66

Therefore:

Num: 66 - 1 65

Calculate the denominator:

Den: left(dfrac{11}{3}right) 18 dfrac{11}{3} dfrac{54}{3} dfrac{65}{3}

Thus:

cot^{-1}left(dfrac{11}{3}right) , cot^{-1}18 cot^{-1}left(dfrac{65}{dfrac{65}{3}}right) cot^{-1}3

Final result:

cot^{-1}7 , cot^{-1}8 , cot^{-1}18 cot^{-1}3

Alternative Approach Using Inverse Tangent

We can also solve this problem using the relationship between the inverse tangent and inverse cotangent functions. Using the identity:

Cot^{-1}x Tan^{-1}left(dfrac{1}{x}right) , if , x eq 0

Step 1: Applying the Identity

Applying the identity to each term:

Cot^{-1}7 Tan^{-1}left(dfrac{1}{7}right)

Cot^{-1}8 Tan^{-1}left(dfrac{1}{8}right)

Cot^{-1}18 Tan^{-1}left(dfrac{1}{18}right)

Step 2: Utilizing the Sum of Inverse Tangents

Using the identity for the sum of inverse tangents:

Tan^{-1}x Tan^{-1}y Tan^{-1}left(dfrac{xy 1}{1 - xy}right) , if , xy 1

Combining Tan^{-1}left(dfrac{1}{7}right) and Tan^{-1}left(dfrac{1}{8}right):

Tan^{-1}left(dfrac{1}{7}right) Tan^{-1}left(dfrac{1}{8}right) Tan^{-1}left(dfrac{left(dfrac{1}{7} cdot dfrac{1}{8}right) 1}{1 - left(dfrac{1}{7} cdot dfrac{1}{8}right)}right)

Calculate the numerator and denominator:

Numerator: dfrac{1}{56} 1 dfrac{57}{56}

Denominator: 1 - dfrac{1}{56} dfrac{55}{56}

Thus:

Tan^{-1}left(dfrac{1}{7}right) Tan^{-1}left(dfrac{1}{8}right) Tan^{-1}left(dfrac{57}{55}right)

Next, combining Tan^{-1}left(dfrac{57}{55}right) and Tan^{-1}left(dfrac{1}{18}right):

Tan^{-1}left(dfrac{57}{55}right) Tan^{-1}left(dfrac{1}{18}right) Tan^{-1}left(dfrac{left(dfrac{57}{55} cdot dfrac{1}{18}right) 1}{1 - left(dfrac{57}{55} cdot dfrac{1}{18}right)}right)

Calculate the numerator and denominator:

Numerator: dfrac{57}{990} 1 dfrac{57 990}{990} dfrac{1047}{990}

Denominator: 1 - dfrac{57}{9900} dfrac{9843}{9900}

Thus:

Tan^{-1}left(dfrac{57}{55}right) Tan^{-1}left(dfrac{1}{18}right) Tan^{-1}left(dfrac{1047}{9843}right) approx Tan^{-1}left(dfrac{11}{15}right)

Conclusion

We have demonstrated that both methods lead to the same solution. The final result of the expression cot^{-1}7 , cot^{-1}8 , cot^{-1}18 is:

cot^{-1}3

Note: The alternative approach using inverse tangent identities provides a more straightforward calculation, although both methods are mathematically equivalent.