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

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,

& \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{aligned}$$ Using the identity: - $$\cos \left( A-B \right)+\cos \left( A+B \right)=2\cos A\cos B$$, we get, $$\begin{aligned} & \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{aligned}$$ Using the property: - $$\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{aligned} & \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{aligned}$$ Now, let us simplify the denominator of L.H.S, i.e. $$\cos A\cos B\cos C$$, so we have, $$\begin{aligned} & \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{aligned}$$ So, substituting the obtained values of numerator and denominator, we get the expression as: - $$\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.