Question

If we have the sum of angles \[A+B+C={{60}^{\circ }}\], then prove that \[\sec A\sec B\sec C+2\sum{\tan A\tan B}=2\].

Answer 1

Hint: Write all the terms by removing summation sign in the second term. Use: - \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[\sec \theta =\dfrac{1}{\cos \theta }\] to simplify the expression. Take L.C.M and convert the product of trigonometric functions into its sum of angles form by using the identity: - \[\cos A\cos B=\dfrac{\cos \left( A-B \right)+\cos \left( A+B \right)}{2}\] and \[\sin A\sin B=\dfrac{\cos \left( A-B \right)-\cos \left( A+B \right)}{2}\]. Finally cancel the common terms to get the answer.

Complete step-by-step solution
Here, we have been provided with the expression: -
L.H.S = \[\sec A\sec B\sec C+2\sum{\tan A\tan B}\]
Removing the summation sign, we get,
L.H.S = \[\sec A\sec B\sec C+2\left( \tan A\tan B+\tan B\tan C+\tan C\tan A \right)\]
Using the conversion, \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[\sec \theta =\dfrac{1}{\cos \theta }\], we get,
L.H.S = \[\dfrac{1}{\cos A\cos B\cos C}+2\left[ \dfrac{\sin A\sin B}{\cos A\cos B}+\dfrac{\sin B\sin C}{\cos B\cos C}+\dfrac{\sin C\sin A}{\cos C\cos A} \right]\]
Taking L.C.M in the above expression, we get,
L.H.S = \[\dfrac{1+2\left( \sin A\sin B\cos C+\sin B\sin C\cos A+\sin A\sin C\cos B \right)}{\cos A\cos B\cos C}\]
Now, \[\sin A\sin B\cos C=\dfrac{1}{2}\left( 2\sin A\sin B \right)\cos C\].
Using the identity: - \[2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)\], we get,
\[\begin{align}
  & \Rightarrow \sin A\sin B\sin C=\dfrac{1}{2}\left[ \cos \left( A-B \right)\cos C-\cos \left( A+B \right)\cos C \right] \\
 & \Rightarrow \sin A\sin B\sin C=\dfrac{1}{2}\times \dfrac{1}{2}\left[ 2\cos \left( A-B \right)\cos C-2\cos \left( A+B \right)\cos C \right] \\
 & \Rightarrow \sin A\sin B\sin C=\dfrac{1}{4}\left[ 2\cos \left( A-B \right)\cos C-2\cos \left( A+B \right)\cos C \right] \\
\end{align}\]
Using the identity: - \[\cos \left( A-B \right)+\cos \left( A+B \right)=2\cos A\cos B\], we get,
\[\begin{align}
  & \Rightarrow \sin A\sin B\cos C=\dfrac{1}{4}\left[ \cos \left( A-B-C \right)+\cos \left( A-B+C \right)-\cos \left( A+B-C \right)-\cos \left( A+B+C \right) \right] \\
 & \Rightarrow \sin A\sin B\cos C=\dfrac{1}{4}\left[ \cos \left[ A-\left( B+C \right) \right]+\cos \left( A+C-B \right)-\cos {{60}^{\circ }}-\cos \left( A+B-C \right) \right] \\
 & \Rightarrow \sin A\sin B\cos C=\dfrac{1}{4}\left[ \cos \left( A-{{60}^{\circ }}+A \right)+\cos \left( {{60}^{\circ }}-B-B \right)-\cos {{60}^{\circ }}-\cos \left( {{60}^{\circ }}-C-C \right) \right] \\
 & \Rightarrow \sin A\sin B\cos C=\dfrac{1}{4}\left[ \cos \left( 2A-{{60}^{\circ }} \right)+\cos \left( {{60}^{\circ }}-2B \right)-\cos {{60}^{\circ }}-\cos \left( {{60}^{\circ }}-2C \right) \right] \\
\end{align}\]
Using the property: - \[\cos \left( A-B \right)=\cos \left( B-A \right)\] and substituting, \[\cos {{60}^{\circ }}=\dfrac{1}{2}\], we get,
\[\Rightarrow \sin A\sin B\cos C=\dfrac{1}{4}\left[ \cos \left( {{60}^{\circ }}-2A \right)+\cos \left( {{60}^{\circ }}-2B \right)-\cos \left( {{60}^{\circ }}-2C \right)-\dfrac{1}{2} \right]\]
On observing the pattern, we have,
\[\begin{align}
  & \Rightarrow \sin B\sin C\cos A=\dfrac{1}{4}\left[ \cos \left( {{60}^{\circ }}-2B \right)+\cos \left( {{60}^{\circ }}-2C \right)-\cos \left( {{60}^{\circ }}-2A \right)-\dfrac{1}{2} \right] \\
 & \Rightarrow \sin C\sin A\cos B=\dfrac{1}{4}\left[ \cos \left( {{60}^{\circ }}-2C \right)+\cos \left( {{60}^{\circ }}-2A \right)-\cos \left( {{60}^{\circ }}-2B \right)+\dfrac{1}{2} \right] \\
\end{align}\]
Now, let us simplify the denominator of L.H.S, i.e. \[\cos A\cos B\cos C\], so we have,
\[\begin{align}
  & \Rightarrow \cos A\cos B\cos C=\dfrac{1}{2}\left( 2\cos A\cos B \right)\cos C \\
 & \Rightarrow \cos A\cos B\cos C=\dfrac{1}{2}\left[ \cos \left( A-B \right)\cos C+\cos \left( A+B \right)\cos C \right] \\
 & \Rightarrow \cos A\cos B\cos C=\dfrac{1}{4}\left[ 2\cos \left( A-B \right)\cos C+\cos \left( A+B \right)\cos C \right] \\
 & \Rightarrow \cos A\cos B\cos C=\dfrac{1}{4}\left[ \cos \left( A-B-C \right)+\cos \left( A-B+C \right)+\cos \left( A+B+C \right)+\cos \left( A+B-C \right) \right] \\
 & \Rightarrow \cos A\cos B\cos C=\dfrac{1}{4}\left[ \cos \left( {{60}^{\circ }}-2A \right)+\cos \left( {{60}^{\circ }}-2B \right)+\cos \left( {{60}^{\circ }}-2C \right)+\dfrac{1}{2} \right] \\
\end{align}\]
So, substituting the obtained values of numerator and denominator, we get the expression as: -
\[\Rightarrow \] L.H.S = \[\dfrac{1+\dfrac{2}{4}\left[ \cos \left( {{60}^{\circ }}-2A \right)+\cos \left( {{60}^{\circ }}-2B \right)+\cos \left( {{60}^{\circ }}-2C \right)-\dfrac{3}{2} \right]}{\dfrac{1}{4}\left[ \cos \left( {{60}^{\circ }}-2A \right)+\cos \left( {{60}^{\circ }}-2B \right)+\cos \left( {{60}^{\circ }}-2C \right)+\dfrac{1}{2} \right]}\]
\[\Rightarrow \] L.H.S = \[\dfrac{4+2\cos \left( {{60}^{\circ }}-2A \right)+2\cos \left( {{60}^{\circ }}-2B \right)+2\cos \left( {{60}^{\circ }}-2C \right)-3}{\cos \left( {{60}^{\circ }}-2A \right)+\cos \left( {{60}^{\circ }}-2B \right)+\cos \left( {{60}^{\circ }}-2C \right)+\dfrac{1}{2}}\]
\[\Rightarrow \] L.H.S = \[2\left[ \dfrac{1+2\left( \cos {{60}^{\circ }}-2A \right)+2\cos \left( {{60}^{\circ }}-2B \right)+2\cos \left( {{60}^{\circ }}-2C \right)}{2\cos \left( {{60}^{\circ }}-2A \right)+2\cos \left( {{60}^{\circ }}-2B \right)+2\cos \left( {{60}^{\circ }}-2C \right)+1} \right]\]
Cancelling the common terms, we get,
\[\Rightarrow \]L.H.S = 2 \[\times \] 1
\[\Rightarrow \] L.H.S = 2 = R.H.S
Hence, proved

Note: It is important to note that while simplifying the expression, \[\sin A\sin B\cos C\]. We have grouped \[\left( \sin A\sin B \right)\] first. One can also group \[\left( \sin A\cos C \right)\] or \[\left( \sin B\cos C \right)\] together and apply the identity: - \[\sin A\sin B=\dfrac{\sin \left( A+B \right)+\sin \left( A-B \right)}{2}\] to simplify. To change sine function into cosine function use the formula: - \[\sin \theta =\cos \left( {{90}^{\circ }}-\theta \right)\]. At last we need to convert all the functions into cosine functions to save time and solve the question in less steps.