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Question: If \({\text{A + B + }}{\text{C = }}{180^o}\), then the value of \(\left( {\cot B + \cot C} \right)\l...

If A + B + C = 180o{\text{A + B + }}{\text{C = }}{180^o}, then the value of (cot⁡B+cot⁡C)(cot⁡C+cot⁡A)(cot⁡A+cot⁡B)\left( {\cot B + \cot C} \right)\left( {\cot C + \cot A} \right)\left( {\cot A + \cot B} \right) will be
(a)\left( a \right) Sec A sec B sec C
(b)\left( b \right) Cosec A cosec B cosec C
(c)\left( c \right) Tan A tan B tan C
(d)\left( d \right) 1

Explanation

Solution

In this particular question use the concept that cot x = (cos x/sin x) and use the concept that sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B, and use the concept that sin⁡(180−x)=sin⁡x\sin \left( {180 - x} \right) = \sin x so use these concepts to reach the solution of the question.

Complete step by step answer:
A + B + C = 180o{\text{A + B + }}{\text{C = }}{180^o}............... (1)
Now we have to find out the value of,
(cot⁡B+cot⁡C)(cot⁡C+cot⁡A)(cot⁡A+cot⁡B)\left( {\cot B + \cot C} \right)\left( {\cot C + \cot A} \right)\left( {\cot A + \cot B} \right)
Now as we know that cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}} so use these properties in the above equation we have,
⇒(cos⁡Bsin⁡B+cos⁡Csin⁡C)(cos⁡Csin⁡C+cos⁡Asin⁡A)(cos⁡Asin⁡A+cos⁡Bsin⁡B)\Rightarrow \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)
Now simplify it we have,
⇒(cos⁡Bsin⁡C+sin⁡Bcos⁡Csin⁡Bsin⁡C)(cos⁡Csin⁡A+sin⁡Ccos⁡Asin⁡Csin⁡A)(cos⁡Asin⁡B+sin⁡Acos⁡Bsin⁡Asin⁡B)\Rightarrow \left( {\dfrac{{\cos B\sin C + \sin B\cos C}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\cos C\sin A + \sin C\cos A}}{{\sin C\sin A}}} \right)\left( {\dfrac{{\cos A\sin B + \sin A\cos B}}{{\sin A\sin B}}} \right)
Now as we know that sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B, so use this property in the above equation we have,
⇒(sin⁡(B+C)sin⁡Bsin⁡C)(sin⁡(A+C)sin⁡Csin⁡A)(sin⁡(A+B)sin⁡Asin⁡B)\Rightarrow \left( {\dfrac{{\sin \left( {B + C} \right)}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\sin \left( {A + C} \right)}}{{\sin C\sin A}}} \right)\left( {\dfrac{{\sin \left( {A + B} \right)}}{{\sin A\sin B}}} \right)
Now from equation (1),
⇒A+B=180o−C\Rightarrow A + B = {180^o} - C
⇒B+C=180o−A\Rightarrow B + C = {180^o} - A
⇒C+A=180o−B\Rightarrow C + A = {180^o} - B
So substitute these values in the above equation we have,
⇒(sin⁡(180o−A)sin⁡Bsin⁡C)(sin⁡(180o−B)sin⁡Csin⁡A)(sin⁡(180o−C)sin⁡Asin⁡B)\Rightarrow \left( {\dfrac{{\sin \left( {{{180}^o} - A} \right)}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\sin \left( {{{180}^o} - B} \right)}}{{\sin C\sin A}}} \right)\left( {\dfrac{{\sin \left( {{{180}^o} - C} \right)}}{{\sin A\sin B}}} \right)
Now as we know that sin⁡(180−x)=sin⁡x\sin \left( {180 - x} \right) = \sin x so we have,
⇒(sin⁡Asin⁡Bsin⁡C)(sin⁡Bsin⁡Csin⁡A)(sin⁡Csin⁡Asin⁡B)\Rightarrow \left( {\dfrac{{\sin A}}{{\sin B\sin C}}} \right)\left( {\dfrac{{\sin B}}{{\sin C\sin A}}} \right)\left( {\dfrac{{\sin C}}{{\sin A\sin B}}} \right)
Now on simplifying we get
⇒(1sin⁡Asin⁡Bsin⁡C)\Rightarrow \left( {\dfrac{1}{{\sin A\sin B\sin C}}} \right)
Now as we know that cos⁡ecx=1sin⁡x\cos ecx = \dfrac{1}{{\sin x}} so we have,
⇒(cot⁡B+cot⁡C)(cot⁡C+cot⁡A)(cot⁡A+cot⁡B)=(1sin⁡Asin⁡Bsin⁡C)=\Rightarrow \left( {\cot B + \cot C} \right)\left( {\cot C + \cot A} \right)\left( {\cot A + \cot B} \right) = \left( {\dfrac{1}{{\sin A\sin B\sin C}}} \right) = Cosec A cosec B cosec C.
So this is the required answer.

So, the correct answer is “Option B”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric properties which are all stated above so use these properties as above and simplify the given equation we will get the required answer. One thing keep in mind that simplify the equation step by step as above simplified otherwise we will get an incorrect answer.