Question

If \[A + B + C = {180^o}\]then \[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right)\]will be
A. \[(secA)(secB)(secC)\]
B. \[(\cos ecA)(\cos ecB)(\cos ecC)\]
C. \[(\tan A)(\tan B)(\tan C)\]
D. \[1\]

Answer 1

Hint:
Here we use the trigonometric ratio \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\] for substitution in the brackets and then solve by taking LCM and apply these properties to simplify the equation. We find the relation between three angles and convert all trigonometric forms to their simplest form by substituting the value of \[\cot \theta \].
* The property \[\sin \left( {180 - \theta } \right) = \sin \theta \]
* Here we use the property to group together \[\cos x\sin y + \sin x\cos y = \sin \left( {x + y} \right)\].

Complete step by step solution:
Given, sum of three angles \[A,B,C\] equal to \[{180^ \circ }\]
i.e. \[\left[ {A + B + C} \right] = {180^o}\]
From the given relation, we can form the following relations between the angles.
Taking one angle to the right hand side of the equation, gives us the sum of the remaining two angles.
\[\left[ {B + C} \right] = {180^o} - \left[ A \right]\]
\[\left[ {A + B} \right] = {180^o} - \left[ C \right]\]
\[\left[ {A + C} \right] = {180^o} - \left[ B \right]\]
Substitute the value of \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\] in the equation \[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right)\] .
\[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right) = \left( {\dfrac{{\cos B}}{{\sin B}} + \dfrac{{\cos C}}{{\sin C}}} \right)\left( {\dfrac{{\cos C}}{{\sin C}} + \dfrac{{\cos A}}{{\sin A}}} \right)\left( {\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\cos B}}{{\sin B}}} \right)\]
Take the LCM inside each parenthesis and convert into a single fraction.
\[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right) = \left( {\dfrac{{\cos B\sin C + \cos C\sin B}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\cos C\sin A + \cos A\sin C}}{{\sin C\sin A}}} \right)\left( {\dfrac{{\cos A\sin B + \cos B\sin A}}{{\sin A\sin B}}} \right)\]
Using the property \[\left( {\cos x\sin y + \sin x\cos y} \right) = \sin \left( {x + y} \right)\]in every parenthesis.
\[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right) = \left( {\dfrac{{\operatorname{Sin} \left( {B + C} \right)}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\operatorname{Sin} \left( {A + C} \right)}}{{\sin C\sin A}}} \right)\left( {\dfrac{{\operatorname{Sin} \left( {A + B} \right)}}{{\sin A\sin B}}} \right)\]
Now, we can use the relations between the angles to solve this equation.
Substitute the value of \[\left[ {B + C} \right] = {180^o} - \left[ A \right]\], \[\left[ {A + B} \right] = {180^o} - \left[ C \right]\] and \[\left[ {A + C} \right] = {180^o} - \left[ B \right]\] in the equation.
\[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right) = \left( {\dfrac{{\operatorname{Sin} \left( {{{180}^o} - A} \right)}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\operatorname{Sin} \left( {{{180}^o} - B} \right)}}{{\sin C\sin A}}} \right)\left( {\dfrac{{\operatorname{Sin} \left( {{{180}^o} - C} \right)}}{{\sin A\sin B}}} \right)\]
Use the property \[\sin \left( {180 - \theta } \right) = \sin \theta \] to simplify.
\[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right) = \left( {\dfrac{{\operatorname{Sin} A}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\operatorname{Sin} B}}{{\sin C\sin A}}} \right)\left( {\dfrac{{\operatorname{Sin} C}}{{\sin A\sin B}}} \right)\]
Cancel out the common factor from numerator and denominator.
\[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right) = \left( {\dfrac{1}{{\operatorname{Sin} A}}} \right)\left( {\dfrac{1}{{\operatorname{Sin} B}}} \right)\left( {\dfrac{1}{{\operatorname{Sin} C}}} \right)\]
As we know \[\dfrac{1}{{\sin \theta }} = \cos ec\theta \], therefore
\[\left( {{\text{cot}}B + {\text{cot}}C} \right)\left( {{\text{cot}}C + {\text{cot}}A} \right)\left( {{\text{cot}}A + {\text{cot}}B} \right) = {\text{cosecA}} \cdot {\text{cosecB}} \cdot {\text{cosecC}}\]

Therefore, Option B is correct.

Note:
 These types of questions always convert all the trigonometric forms in \[\sin \] and \[\cos \] form which makes it easier to solve. Common mistake done by the students is when they start solving these questions directly by multiplying the brackets which gives a very complex equation to solve. Students should avoid solving these kinds of questions by direct multiplication as it will give very complex calculations where we might not be able to use our trigonometric identities to solve the equation.
Other important trigonometric forms which may help in these type of questions are
1) \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
2) \[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
3) \[1 + {\cot ^2}\theta = \cos e{c^2}\theta \]
4) \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]