Question

If two trigonometric equations are given as \[\sin A\sin B\sin C=p\] and \[\cos A\cos B\cos C=q\] where A,B and C are the angles of \[\Delta ABC\], then \[\tan A\tan B+\tan B\tan C+\tan C\tan A\] is equal to
\[\begin{align}
  & A)\dfrac{p}{q} \\
 & B)\dfrac{q+1}{q} \\
 & C)\dfrac{p+1}{p} \\
 & D)\dfrac{q+1}{p} \\
\end{align}\]

Answer 1

Hint: From the question, we were given that \[\sin A\sin B\sin C=p\] and \[\cos A\cos B\cos C=q\]. Let us assume \[\sin A\sin B\sin C=p\] as equation (1). Now, let us assume \[\cos A\cos B\cos C=q\] as equation (2).We know that if A, B and C are the angles of \[\Delta ABC\], then \[\cos (A+B+C)=-1\]. Let us assume this as equation (3). We also know that \[\cos (A+B+C)=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C\]. Let us assume this as equation (4). Now we will substitute equation (3) in equation (4). Now we should simplify. In this way, we can get the value of \[\tan A\tan B+\tan B\tan C+\tan C\tan A\].

Complete step by step answer:
From the question, we were given that \[\sin A\sin B\sin C=p\] and \[\cos A\cos B\cos C=q\].
Let us consider
\[\sin A\sin B\sin C=p.....(1)\]
\[\cos A\cos B\cos C=q.....(2)\]
We know that if A, B and C are the angles of \[\Delta ABC\], then \[\cos (A+B+C)=-1\].
Let us consider
\[\cos (A+B+C)=-1......(3)\]
 We also know that \[\cos (A+B+C)=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C\]
Let us consider
\[\cos (A+B+C)=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C.....(4)\]
Now let us substitute equation (3) in equation (4), then we get
Let us take \[\cos A\cos B\cos C\]as common.
\[\Rightarrow -1=\cos A\cos B\cos C\left( 1-\dfrac{\sin A\sin B\cos C}{\cos A\cos B\cos C}-\dfrac{\sin A\cos B\sin C}{\cos A\cos B\cos C}-\dfrac{\cos A\sin B\sin C}{\cos A\cos B\cos C} \right)\]
We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\].
\[\Rightarrow -1=\cos A\cos B\cos C\left( 1-\tan A\tan B-\tan B\operatorname{tanC}-\tan C\tan A \right).....(5)\]
Now let us substitute equation (2) in equation (5), then we get
\[\Rightarrow -1=q\left( 1-\tan A\tan B-\tan B\operatorname{tanC}-\tan C\tan A \right)\]
Now we will apply cross multiplication, then we get
\[\Rightarrow 1-\tan A\tan B-\tan B\operatorname{tanC}-\tan C\tan A=\dfrac{-1}{q}\]
\[\Rightarrow \tan A\tan B+\tan B\operatorname{tanC}+\tan C\tan A=1+\dfrac{1}{q}\]
\[\Rightarrow \tan A\tan B+\tan B\operatorname{tanC}+\tan C\tan A=\dfrac{1+q}{q}\]

So, the correct answer is “Option B”.

Note: Students may have a misconception that \[sin(A+B+C)=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C\]. But we know that \[\cos (A+B+C)=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C\]. So, these misconceptions should be avoided. Otherwise, it may result in wrong answers. Students should also be careful at the calculation part of the solution. If a small mistake is done, then the whole solution may go wrong. So, the calculation part should be done in a careful manner. Students should also avoid silly mistakes while solving this problem. These silly mistakes may vary a huge change in result.