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Question: If \({\cot ^{ - 1}}\alpha + {\cot ^{ - 1}}\beta = {\cot ^{ - 1}}x\), then the value of x is: (A) ...

If cot⁡−1α+cot⁡−1β=cot⁡−1x{\cot ^{ - 1}}\alpha + {\cot ^{ - 1}}\beta = {\cot ^{ - 1}}x, then the value of x is:
(A) α+β\alpha + \beta
(B) α−β\alpha - \beta
(C) 1+αβα+β\dfrac{{1 + \alpha \beta }}{{\alpha + \beta }}
(D) αβ−1α+β\dfrac{{\alpha \beta - 1}}{{\alpha + \beta }}

Explanation

Solution

Hint : In the given question, we are provided with an equation involving the inverse trigonometric functions such as cotangent inverse function and we have to find the value of x. First, we take cotangent on both sides of the equation and then simplify the equation to find the value of variable x. We will use the compound angle formula cot⁡(A+B)=cot⁡Acot⁡B−1cot⁡A+cot⁡B\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}. Basic algebraic rules and trigonometric identities are to be kept in mind while simplifying the given problem and proving the result given to us.

Complete step-by-step answer :
So, we have, cot⁡−1α+cot⁡−1β=cot⁡−1x{\cot ^{ - 1}}\alpha + {\cot ^{ - 1}}\beta = {\cot ^{ - 1}}x.
Taking cotangent on both the sides of equation, we get,
⇒cot⁡(cot⁡−1α+cot⁡−1β)=cot⁡(cot⁡−1x)\Rightarrow \cot \left( {{{\cot }^{ - 1}}\alpha + {{\cot }^{ - 1}}\beta } \right) = \cot \left( {{{\cot }^{ - 1}}x} \right)
Now, we know that the value of the expression cot⁡(cot⁡−1x)=x\cot \left( {{{\cot }^{ - 1}}x} \right) = x. So, we get,
⇒cot⁡(cot⁡−1α+cot⁡−1β)=x\Rightarrow \cot \left( {{{\cot }^{ - 1}}\alpha + {{\cot }^{ - 1}}\beta } \right) = x
Now, we use the trigonometric compound formula for cotangent cot⁡(A+B)=cot⁡Acot⁡B−1cot⁡A+cot⁡B\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}. So, we get,
⇒cot⁡(cot⁡−1α)×cot⁡(cot⁡−1β)−1cot⁡(cot⁡−1α)+cot⁡(cot⁡−1β)=x\Rightarrow \dfrac{{\cot \left( {{{\cot }^{ - 1}}\alpha } \right) \times \cot \left( {{{\cot }^{ - 1}}\beta } \right) - 1}}{{\cot \left( {{{\cot }^{ - 1}}\alpha } \right) + \cot \left( {{{\cot }^{ - 1}}\beta } \right)}} = x
Again using the formula cot⁡(cot⁡−1x)=x\cot \left( {{{\cot }^{ - 1}}x} \right) = x and simplifying the equation, we get,
⇒α×β−1α+β=x\Rightarrow \dfrac{{\alpha \times \beta - 1}}{{\alpha + \beta }} = x
Changing the sides of equation and simplifying further, we get,
⇒x=αβ−1α+β\Rightarrow x = \dfrac{{\alpha \beta - 1}}{{\alpha + \beta }}
So, the value of x is αβ−1α+β\dfrac{{\alpha \beta - 1}}{{\alpha + \beta }}.
So, the correct answer is “Option (D)”.

Note : For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. We must remember the compound angle formula of cotangent function in order to solve the problem. One must take care while handling the calculations.