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Question: If \[\dfrac{{\cos (A - B)}}{{\cos (A + B)}} + \dfrac{{\cos (C + D)}}{{\cos (C - D)}} = 0\] then A...

If cos⁡(A−B)cos⁡(A+B)+cos⁡(C+D)cos⁡(C−D)=0\dfrac{{\cos (A - B)}}{{\cos (A + B)}} + \dfrac{{\cos (C + D)}}{{\cos (C - D)}} = 0 then
A.tan⁡Atan⁡B=tan⁡Ctan⁡D\tan A\tan B = \tan C\tan D
B.tan⁡Atan⁡B=−tan⁡Ctan⁡D\tan A\tan B = - \tan C\tan D
C.tan⁡Atan⁡Btan⁡Ctan⁡D=−1\tan A\tan B\tan C\tan D = - 1
D.None of these

Explanation

Solution

Hint : In mathematics the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. The trigonometric functions most widely used in modern mathematics are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used. Each of these six trigonometric functions has a corresponding inverse function called the inverse trigonometric function . The oldest definitions of trigonometric functions, related to right-angle triangles, define them only for acute angles.
Formulas used in the solution part:
cos⁡(C−D)=cos⁡Ccos⁡D+sin⁡Csin⁡D\cos (C - D) = \cos C\cos D + \sin C\sin D
cos⁡(C+D)=cos⁡Ccos⁡D−sin⁡Csin⁡D\cos (C + D) = \cos C\cos D - \sin C\sin D

Complete step-by-step answer :
We have the equation cos⁡(A−B)cos⁡(A+B)+cos⁡(C+D)cos⁡(C−D)=0\dfrac{{\cos (A - B)}}{{\cos (A + B)}} + \dfrac{{\cos (C + D)}}{{\cos (C - D)}} = 0
Using the identities stated above we get
cos⁡Acos⁡B+sin⁡Asin⁡Bcos⁡Acos⁡B−sin⁡Asin⁡B+cos⁡Ccos⁡D−sin⁡Csin⁡Dcos⁡Ccos⁡D+sin⁡Csin⁡D=0\dfrac{{\cos A\cos B + \sin A\sin B}}{{\cos A\cos B - \sin A\sin B}} + \dfrac{{\cos C\cos D - \sin C\sin D}}{{\cos C\cos D + \sin C\sin D}} = 0
Now taking all the cosine terms as common we get the following equation ,
1+tan⁡Atan⁡B1−tan⁡Atan⁡B+1−tan⁡Ctan⁡D1+tan⁡Ctan⁡D=0\dfrac{{1 + \tan A\tan B}}{{1 - \tan A\tan B}} + \dfrac{{1 - \tan C\tan D}}{{1 + \tan C\tan D}} = 0
Now on taking the LCM ( Least Common Multiple) we get ,
(1+tan⁡Atan⁡B)(1+tan⁡Ctan⁡D)+(1−tan⁡Ctan⁡D)(1−tan⁡Atan⁡B)(1−tan⁡Atan⁡B)(1+tan⁡Ctan⁡D)=0\dfrac{{\left( {1 + \tan A\tan B} \right)\left( {1 + \tan C\tan D} \right) + \left( {1 - \tan C\tan D} \right)\left( {1 - \tan A\tan B} \right)}}{{\left( {1 - \tan A\tan B} \right)\left( {1 + \tan C\tan D} \right)}} = 0
Which on further simplification becomes ,
(1+tan⁡Atan⁡B)(1+tan⁡Ctan⁡D)+(1−tan⁡Ctan⁡D)(1−tan⁡Atan⁡B)=0\left( {1 + \tan A\tan B} \right)\left( {1 + \tan C\tan D} \right) + \left( {1 - \tan C\tan D} \right)\left( {1 - \tan A\tan B} \right) = 0
On simplifying this equation further we get ,
2+2(tan⁡Atan⁡Btan⁡Ctan⁡D)=02 + 2\left( {\tan A\tan B\tan C\tan D} \right) = 0
Hence finally we get the relation ,
tan⁡Atan⁡Btan⁡Ctan⁡D=−1\tan A\tan B\tan C\tan D = - 1
Therefore option (3) is the correct answer.
So, the correct answer is “Option 3”.

Note: The trigonometric functions most widely used in modern mathematics are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used. Each of these six trigonometric functions has a corresponding inverse function called the inverse trigonometric function . keep all the trigonometric identities in mind and use them carefully.