Question

In any $\Delta ABC$, prove the following:
$\dfrac{b\sec B+c\sec C}{\tan B+\tan C}=\dfrac{c\sec C+a\sec A}{\tan C+\tan A}=\dfrac{a\sec A+b\sec B}{\tan A+\tan B}$.

Answer 1

Hint: We use the properties of triangles involving the values of sides, angles and the circum-radius R of $\Delta ABC$. We have to show that the three expressions of trigo are the same. We find their individual value using the properties. All of them are equal to 2R.

Complete step by step answer:
To prove the given equation, we are going to use some properties of triangles involving trigonometric ratios of the angles and the sides.
For $\Delta ABC$, the opposite sides of the angles $\angle A,\angle B,\angle C$ are $a,b,c$ respectively.

So, we have the identities $\sec A=\dfrac{2bc}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}$. Similarly, we have $\sec B=\dfrac{2ac}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}$ and $\sec C=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}$.
We also have $\tan A=\dfrac{abc}{R\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)},\tan B=\dfrac{abc}{R\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)},\tan C=\dfrac{abc}{R\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}$.
Here R defines the circum-radius of $\Delta ABC$.
Now we place the values in the equations.
For $\dfrac{b\sec B+c\sec C}{\tan B+\tan C}$, we get $\dfrac{b\sec B+c\sec C}{\tan B+\tan C}=\dfrac{b\left( \dfrac{2ac}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}} \right)+c\left( \dfrac{2ab}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}} \right)}{\dfrac{abc}{R\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)}+\dfrac{abc}{R\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}}$.
We solve to get $\dfrac{b\sec B+c\sec C}{\tan B+\tan C}=\dfrac{2abc\left( \dfrac{1}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}+\dfrac{1}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}} \right)}{\dfrac{abc}{R}\left( \dfrac{1}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}+\dfrac{1}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}} \right)}=2R$.
For $\dfrac{c\sec C+a\sec A}{\tan C+\tan A}$, we get $\dfrac{c\sec C+a\sec A}{\tan C+\tan A}=\dfrac{c\left( \dfrac{2ab}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}} \right)+a\left( \dfrac{2bc}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}} \right)}{\dfrac{abc}{R\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}+\dfrac{abc}{R\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}}$.
We solve to get \[\dfrac{c\sec C+a\sec A}{\tan C+\tan A}=\dfrac{2abc\left( \dfrac{1}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}+\dfrac{1}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}} \right)}{\dfrac{abc}{R}\left( \dfrac{1}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}+\dfrac{1}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}} \right)}=2R\].
For $\dfrac{a\sec A+b\sec B}{\tan A+\tan B}$, we get $\dfrac{a\sec A+b\sec B}{\tan A+\tan B}=\dfrac{a\left( \dfrac{2bc}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}} \right)+b\left( \dfrac{2ac}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}} \right)}{\dfrac{abc}{R\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}+\dfrac{abc}{R\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)}}$.
We solve to get $\dfrac{a\sec A+b\sec B}{\tan A+\tan B}=\dfrac{2abc\left( \dfrac{1}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}+\dfrac{1}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}} \right)}{\dfrac{abc}{R}\left( \dfrac{1}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}+\dfrac{1}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}} \right)}=2R$.
So, $\dfrac{b\sec B+c\sec C}{\tan B+\tan C}=\dfrac{c\sec C+a\sec A}{\tan C+\tan A}=\dfrac{a\sec A+b\sec B}{\tan A+\tan B}=2R$. Thus proved.

Note: In the final step of every solution finding we eliminated terms like $a,b,c$ and $\dfrac{1}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}+\dfrac{1}{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}$. That’s possible only because all the values of the expressions are positive. Otherwise it wouldn’t have been possible. The formulas of $\sec \theta $ are coming from the inverse theorem of $\cos \theta $. For example: $\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$. We took both side inverse to get $\dfrac{1}{\cos A}=\dfrac{1}{\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}}\Rightarrow \sec A=\dfrac{2bc}{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}$. Same goes for other angles too.