Question

If $\dfrac{\cos \left( A-B \right)}{\cos \left( A+B \right)}+\dfrac{\cos \left( C+D \right)}{\cos \left( C-D \right)}=0$, then tan A tan B tan C =
(a) tan D
(b) cot D
(c) – tan D
(d) – cot D

Answer 1

Hint: First of all expand all the terms by using the general formulas. Then try to convert sin, and cos into terms of tan. Now, take the least common multiple and try to generate the term tan A tan B tan C. Cancel all the remaining terms with opposite signs. The final relation of tangents A, B, C, D is the required result.

Complete step-by-step answer:
The given equation in the question can be written as,
$\dfrac{\cos \left( A-B \right)}{\cos \left( A+B \right)}+\dfrac{\cos \left( C+D \right)}{\cos \left( C-D \right)}=0$
By basic knowledge of trigonometry, we know the formulas
$\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$
$\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y$
By substituting them into the equation, we get it as,
$\dfrac{\cos A\cos B+\sin A\sin B}{\cos A\cos B-\sin A\sin B}+\dfrac{\cos C\cos D-\sin C\sin D}{\cos C\cos D+\sin C\sin D}=0$
By dividing with cos A cos B, cos C cos D on the numerator and denominator of the first term and second term respectively, we get,
$\dfrac{\dfrac{\cos A\cos B+\sin A\sin B}{\cos A\cos B}}{\dfrac{\cos A\cos B-\sin A\sin B}{\cos A\cos B}}+\dfrac{\dfrac{\cos C\cos D-\sin C\sin D}{\cos C\cos D}}{\dfrac{\cos C\cos D+\sin C\sin D}{\cos C\cos D}}=0$
By expanding each fraction, we can write the equation as
$\dfrac{1+\tan A\tan B}{1-\tan A\tan B}+\dfrac{1-\tan C\tan D}{1+\tan C\tan D}=0$
By taking the least common multiple, we get the equation as:
$\dfrac{\left( 1+\tan A\tan B \right)\left( 1+\tan C\tan D \right)+\left( 1-\tan C\tan D \right)\left( 1-\tan A\tan B \right)}{\left( 1-\tan A\tan B \right)\left( 1+\tan C\tan D \right)}=0$
By multiplying with the denominator on both the sides, we get it as:
$\left( 1+\tan A\tan B \right)\left( 1+\tan C\tan D \right)+\left( 1-\tan C\tan D \right)\left( 1-\tan A\tan B \right)=0$
By using distributive law on the first product, we can write
$1+\tan A\tan B+\tan C\tan D+\tan A\tan B\tan C\tan D+\left( 1-\tan C\tan D \right)\left( 1-\tan A\tan B \right)=0$
By expanding the second product in the same way, we get,
$1+\tan A\tan B+\tan C\tan D+\tan A\tan B\tan C\tan D+1-\tan A\tan B-\tan C\tan D+\tan A\tan B\tan C\tan D=0$
By grouping the common terms, we get,
$\left( 1+1 \right)+\left( \tan A\tan B \right)\left( 1-1 \right)+\tan C\tan D\left( 1-1 \right)+\tan A\tan B\tan C\tan D\left( 1+1 \right)=0$
By simplifying the above equation, we get the equation as,
$2+\tan A\tan B\left( 0 \right)+\tan C\tan D\left( 0 \right)+\tan A\tan B\tan C\tan D\left( 2 \right)=0$
By further simplifying, we get the equation as,
$2\tan A\tan B\tan C\tan D+2=0$
By subtracting and dividing by 2 on both the sides, we get,
$\tan A\tan B\tan C\tan D=-1$
By dividing by tan D on both the sides, we get its value as,
$\tan A\tan B\tan C=-\cot D$
Hence, option (d) is the right answer.

Note: Be careful while substituting the formula of cos (A – B) cos (A + B) as students confuse between +, - and get the reverse answer. So, do it carefully as we need the term tan A tan B tan C. So, that’s why we divide cosine products in fraction and simplify taking the least common multiple. Simplify each step carefully to get the desired result in the question.