Let a, b, c be positive real numbers....

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1981Moderate

Let $a, b, c$ be positive real numbers. Let $θ = tan^{- 1} \frac{a ( a + b + c )}{b c} + tan^{- 1} \frac{b ( a + b + c )}{c a} + tan^{- 1} \frac{c ( a + b + c )}{ab}$ . Then $tan θ =$ .........

Answer:0

Visualized Solution (Hindi)

Substitution of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a + b + c

Let $a + b + c = u$
Given expression: $θ = tan^{- 1} \frac{a u}{b c} + tan^{- 1} \frac{b u}{c a} + tan^{- 1} \frac{c u}{ab}$
Note: Since $a, b, c > 0$ , we have $u > 0$ .

Identifying the Variables .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x, y, z

Let $x = \frac{a u}{b c}$ , $y = \frac{b u}{c a}$ , and $z = \frac{c u}{ab}$
The expression becomes: $θ = tan^{- 1} x + tan^{- 1} y + tan^{- 1} z$

Checking the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x y

Calculate the product $x y$ : $x y = \frac{a u}{b c} \cdot \frac{b u}{c a}$
$x y = \frac{ab u ^{2}}{ab c ^{2}} = \frac{u}{c}$
Since $u = a + b + c$ , then $\frac{u}{c} = \frac{a + b + c}{c} = 1 + \frac{a + b}{c}$
As $a, b, c > 0$ , we conclude $x y > 1$ .

Applying the Adjusted Formula

Since $x, y > 0$ and $x y > 1$ , we use the formula:
$tan^{- 1} x + tan^{- 1} y = π + tan^{- 1} (\frac{x + y}{1 - x y})$
So, $θ = π + tan^{- 1} (\frac{x + y}{1 - x y}) + tan^{- 1} z$

Simplifying the Numerator .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x + y

Numerator: $x + y = \frac{a u}{b c} + \frac{b u}{c a}$
Factor out $u$ : $x + y = u (\frac{a}{b c} + \frac{b}{c a})$
Common denominator $ab c$ : $x + y = u (\frac{a + b}{ab c})$

Simplifying the Denominator .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 - x y

Denominator: $1 - x y = 1 - \frac{u}{c}$
Substitute $u = a + b + c$ : $1 - x y = 1 - \frac{a + b + c}{c}$
$1 - x y = \frac{c - ( a + b + c )}{c} = \frac{- ( a + b )}{c}$

Combining Numerator and Denominator

$\frac{x + y}{1 - x y} = \frac{\frac{( a + b ) u}{ab c}}{\frac{- ( a + b )}{c}}$
Cancel $(a + b)$ : $\frac{x + y}{1 - x y} = \frac{u}{ab c} \cdot \frac{c}{- 1}$
Simplify: $\frac{x + y}{1 - x y} = - \frac{c u}{ab c} = - \frac{c ^{2} u}{ab c} = - \frac{c u}{ab}$

Relating the Result to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Recall $z = \frac{c u}{ab}$
Therefore, $\frac{x + y}{1 - x y} = - z$
The equation for $θ$ becomes: $θ = π + tan^{- 1} (- z) + tan^{- 1} z$

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Using the property $tan^{- 1} (- z) = - tan^{- 1} z$ :
$θ = π - tan^{- 1} z + tan^{- 1} z$
$θ = π$

Final Answer for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan θ

We need to find $tan θ$
Since $θ = π$ , then $tan θ = tan π$
Final result: $tan θ = 0$

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