If $ \cos x + \cos y = \dfrac{4}{5},\cos x - \cos y = \dfrac{2}{7}. $ The value of $ 14\tan \left( {\dfrac{{x - y}}{2}} \right) + 5\cot \left( {\dfrac{{x + y}}{2}} \right) $ is
A. $ 0 $
B. $ \dfrac{1}{4} $
C. $ \dfrac{5}{4} $
D. $ \dfrac{3}{4} $
Answer
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Hint: First of all we will use identity for the difference of two cosine angles and the sum of two cosine angles and then find correlation to get the equation framed in tangent and cot and then simplify for the resultant required value.
Complete step-by-step answer:
Take the given expression:
$ \cos x + \cos y = \dfrac{4}{5} $
By applying the identity for the sum of two cosine angles
$ 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) = \dfrac{4}{5} $
Term multiplicative at one side if moved to the opposite side, then it goes in the denominator. Also, common factors in the numerator and the denominator cancels each other.
$ \cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) = \dfrac{2}{5} $ ….. (A)
Similarly, take the second given equation –
$ \cos x - \cos y = \dfrac{2}{7}. $
Apply identity for the difference of two cosines,
$ - 2\sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right) = \dfrac{2}{7} $
Common multiples from both the sides of the equation cancel each other and therefore remove from both the sides of the equation.
$ - \sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right) = \dfrac{1}{7} $ ….. (B)
Divide equation (B) with the equation (A)
$ \dfrac{{ - \sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)}} = \dfrac{{\dfrac{1}{7}}}{{\dfrac{2}{5}}} $
We know that tangent is expressed as the ratio of sine upon cosine and cot angle is expressed as the cosine upon sine. Also, denominator’s denominator goes to the numerator.
\[\dfrac{{ - \tan \left( {\dfrac{{x + y}}{2}} \right)}}{{\cot \left( {\dfrac{{x - y}}{2}} \right)}} = \dfrac{5}{{7 \times 2}}\]
Simplify the above expression finding the product in the denominator on the right hand side of the equation.
\[\dfrac{{ - \tan \left( {\dfrac{{x + y}}{2}} \right)}}{{\cot \left( {\dfrac{{x - y}}{2}} \right)}} = \dfrac{5}{{14}}\]
Perform cross multiplication in the above expression where the numerator of one side is multiplied with the denominator of the opposite side and vice versa.
\[ - 14\tan \left( {\dfrac{{x + y}}{2}} \right) = 5\cot \left( {\dfrac{{x - y}}{2}} \right)\]
Move the term tangent on the opposite side, when any term is moved to the opposite its sign changes.
\[0 = 5\cot \left( {\dfrac{{x - y}}{2}} \right) + 14\tan \left( {\dfrac{{x + y}}{2}} \right)\]
The above equation can be re-written as –
\[5\cot \left( {\dfrac{{x - y}}{2}} \right) + 14\tan \left( {\dfrac{{x + y}}{2}} \right) = 0\]
From the given multiple choices, option A is the correct answer.
So, the correct answer is “Option A”.
Note: Be good in different identities to simplify the given expression in the required format and also remember basic mathematics operators to get the required value. Be clear with the identity for difference of two cosine angles and the sum of two cosine angles.
Complete step-by-step answer:
Take the given expression:
$ \cos x + \cos y = \dfrac{4}{5} $
By applying the identity for the sum of two cosine angles
$ 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) = \dfrac{4}{5} $
Term multiplicative at one side if moved to the opposite side, then it goes in the denominator. Also, common factors in the numerator and the denominator cancels each other.
$ \cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) = \dfrac{2}{5} $ ….. (A)
Similarly, take the second given equation –
$ \cos x - \cos y = \dfrac{2}{7}. $
Apply identity for the difference of two cosines,
$ - 2\sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right) = \dfrac{2}{7} $
Common multiples from both the sides of the equation cancel each other and therefore remove from both the sides of the equation.
$ - \sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right) = \dfrac{1}{7} $ ….. (B)
Divide equation (B) with the equation (A)
$ \dfrac{{ - \sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)}} = \dfrac{{\dfrac{1}{7}}}{{\dfrac{2}{5}}} $
We know that tangent is expressed as the ratio of sine upon cosine and cot angle is expressed as the cosine upon sine. Also, denominator’s denominator goes to the numerator.
\[\dfrac{{ - \tan \left( {\dfrac{{x + y}}{2}} \right)}}{{\cot \left( {\dfrac{{x - y}}{2}} \right)}} = \dfrac{5}{{7 \times 2}}\]
Simplify the above expression finding the product in the denominator on the right hand side of the equation.
\[\dfrac{{ - \tan \left( {\dfrac{{x + y}}{2}} \right)}}{{\cot \left( {\dfrac{{x - y}}{2}} \right)}} = \dfrac{5}{{14}}\]
Perform cross multiplication in the above expression where the numerator of one side is multiplied with the denominator of the opposite side and vice versa.
\[ - 14\tan \left( {\dfrac{{x + y}}{2}} \right) = 5\cot \left( {\dfrac{{x - y}}{2}} \right)\]
Move the term tangent on the opposite side, when any term is moved to the opposite its sign changes.
\[0 = 5\cot \left( {\dfrac{{x - y}}{2}} \right) + 14\tan \left( {\dfrac{{x + y}}{2}} \right)\]
The above equation can be re-written as –
\[5\cot \left( {\dfrac{{x - y}}{2}} \right) + 14\tan \left( {\dfrac{{x + y}}{2}} \right) = 0\]
From the given multiple choices, option A is the correct answer.
So, the correct answer is “Option A”.
Note: Be good in different identities to simplify the given expression in the required format and also remember basic mathematics operators to get the required value. Be clear with the identity for difference of two cosine angles and the sum of two cosine angles.
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