Solveeit Logo

Question

Question: The formula of \[\cot \left( {A - B} \right)\] is?...

The formula of cot(AB)\cot \left( {A - B} \right) is?

Explanation

Solution

Hint : Here the question is related to trigonometry, we have to find the formula of cotangent or cot difference identity or formula. This can be find by using tangent difference identity tan(AB)=tanAtanB1+tanAtanB\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \cdot \tan B}} , and further simplification using a reciprocal definition of trigonometric function we get the required solution.

Complete step by step solution:
Trigonometric ratios: Some ratios of the sides of a right-angle triangle with respect to its acute angle called trigonometric ratios of the angle.
The ratios defined are abbreviated as sin A, cos A, tan A, csc A or cosec A, sec A and cot A
These functions are defined as the reciprocal of the standard trigonometric functions: sine, cosine, and tangent, and hence they are called the reciprocal trigonometric functions.
The reciprocal trigonometric functions are: 1sinA=cosecA\dfrac{1}{{\sin A}} = \cos ec\,A , 1cosA=secA\dfrac{1}{{\cos A}} = sec\,A and 1tanA=cotA\dfrac{1}{{\tan A}} = \cot \,A .
Consider the given question:
We need to find the formula of
cot(AB)\Rightarrow \cot \left( {A - B} \right)
Now, using the tangent or tan sum identity i.e., tan(AB)=tanAtanB1+tanAtanB\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \cdot \tan B}} .
Take reciprocal of tan sum identity, then
1tan(AB)=1+tanAtanBtanAtanB\Rightarrow \dfrac{1}{{\tan \left( {A - B} \right)}} = \dfrac{{1 + \tan A \cdot \tan B}}{{\tan A - \tan B}}
By using a Reciprocal trigonometric function 1tanA=cotA\dfrac{1}{{\tan A}} = \cot \,A or 1cotA=tanA\dfrac{1}{{\cot A}} = \tan A , then we have
cot(AB)=1+1cotA1cotB1cotA1cotB\Rightarrow \cot \left( {A - B} \right) = \dfrac{{1 + \dfrac{1}{{\cot A}} \cdot \dfrac{1}{{\cot B}}}}{{\dfrac{1}{{\cot A}} - \dfrac{1}{{\cot B}}}}
Take cotAcotB\cot A\cot B as LCM in both numerator and denominator of RHS, then
cot(AB)=cotAcotB+1cotAcotBcotBcotAcotAcotB\Rightarrow \cot \left( {A - B} \right) = \dfrac{{\dfrac{{\cot A\cot B + 1}}{{\cot A\cot B}}}}{{\dfrac{{\cot B - \cot A}}{{\cot A\cot B}}}}
On cancelling the like terms in both numerator and denominator of RHS, then we get
cot(AB)=cotAcotB+1cotBcotA\Rightarrow \cot \left( {A - B} \right) = \dfrac{{\cot A\cot B + 1}}{{\cot B - \cot A}}
Hence, the required formula of cot(AB)=cotAcotB+1cotBcotA\cot \left( {A - B} \right) = \dfrac{{\cot A\cot B + 1}}{{\cot B - \cot A}} .
So, the correct answer is “ cot(AB)=cotAcotB+1cotBcotA\cot \left( {A - B} \right) = \dfrac{{\cot A\cot B + 1}}{{\cot B - \cot A}} .”.

Note : The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. We must know about the trigonometry definitions, identities and formulas which are involving the trigonometry ratios, while solving the trigonometry based questions.