Question

Prove that in triangle ABC: $\left( {\tan A + \tan B + \tan C} \right)\left( {\cot A + \cot B + \cot C} \right) = 1 + \sec A\sec B\sec C$

Answer 1

Hint – In this question use the angle sum property of triangle to write tan of angle A+B in terms of tan C. Then use the direct formula of $\tan \left( {A + B} \right)$ that is $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$. Simplify the L.H.S part by writing cot x in terms of $\cot x = \dfrac{1}{{\tan x}}$. This will help getting the right hand side.

Complete step-by-step answer:
As we know in triangle the sum of all angle is ${180^o} = \pi $
$ \Rightarrow A + B + C = \pi $................... (1)
Now take C to R.H.S and take tan on both sides we have,
$ \Rightarrow \tan \left( {A + B} \right) = \tan \left( {\pi - C} \right)$
Now apply tan rule we have,
$ \Rightarrow \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = - \tan C$, $\left[ {\because \tan \left( {\pi - C} \right) = - \tan C} \right]$
Now simplify it we have,
$ \Rightarrow \tan A + \tan B = - \tan C\left( {1 - \tan A\tan B} \right)$
$ \Rightarrow \tan A + \tan B = - \tan C + \tan A\tan B\tan C$
$ \Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$..................... (2)
Now take L.H.S of the given equation we have,
$ \Rightarrow \left( {\tan A + \tan B + \tan C} \right)\left( {\cot A + \cot B + \cot C} \right)$
Now as we know $\cot x = \dfrac{1}{{\tan x}}$ so use this property in above equation we have,
$ \Rightarrow \left( {\tan A + \tan B + \tan C} \right)\left( {\dfrac{1}{{\tan A}} + \dfrac{1}{{\tan B}} + \dfrac{1}{{\tan C}}} \right)$
Now simplify this equation we have,
$ \Rightarrow \left( {\tan A + \tan B + \tan C} \right)\left( {\dfrac{{\tan B\tan C + \tan A\tan C + \tan A\tan B}}{{\tan A\tan B\tan C}}} \right)$
Now substitute the value from equation (2) we have,
$ \Rightarrow \left( {\tan A\tan B\tan C} \right)\left( {\dfrac{{\tan B\tan C + \tan A\tan C + \tan A\tan B}}{{\tan A\tan B\tan C}}} \right)$
$ \Rightarrow \tan B\tan C + \tan A\tan C + \tan A\tan B$
Now as we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ so use this property in above equation we have,
\[ \Rightarrow \dfrac{{\sin B}}{{\cos B}}\dfrac{{\sin C}}{{\cos C}} + \dfrac{{\sin A}}{{\cos A}}\dfrac{{\sin C}}{{\cos C}} + \dfrac{{\sin A}}{{\cos A}}\dfrac{{\sin B}}{{\cos B}}\]
Now take cos A cos B cos C as L.C.M we have,
\[ \Rightarrow \dfrac{{\cos A\sin B\sin C + \cos B\sin A\sin C + \cos C\sin A\sin B}}{{\cos A\cos B\cos C}}\]
Now take sin C common from the numerator first and second term we have,
\[ \Rightarrow \dfrac{{\sin C\left( {\cos A\sin B + \cos B\sin A} \right) + \cos C\sin A\sin B}}{{\cos A\cos B\cos C}}\]
Now as we know that $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ so we have,
\[ \Rightarrow \dfrac{{\sin C\sin \left( {A + B} \right) + \cos C\sin A\sin B}}{{\cos A\cos B\cos C}}\]
Now from equation (1) we have,
\[ \Rightarrow \dfrac{{\sin C\sin \left( {\pi - C} \right) + \cos C\sin A\sin B}}{{\cos A\cos B\cos C}}\]
\[ \Rightarrow \dfrac{{{{\sin }^2}C + \cos C\sin A\sin B}}{{\cos A\cos B\cos C}}\], $\left[ {\because \sin \left( {\pi - C} \right) = \sin C} \right]$
Now use ${\sin ^2}C = 1 - {\cos ^2}C$ so we have,
\[ \Rightarrow \dfrac{{1 - {{\cos }^2}C + \cos C\sin A\sin B}}{{\cos A\cos B\cos C}}\]
\[ \Rightarrow \dfrac{1}{{\cos A\cos B\cos C}} - \dfrac{{\cos C\left( {\cos C - \sin A\sin B} \right)}}{{\cos A\cos B\cos C}}\]
\[ \Rightarrow \dfrac{1}{{\cos A\cos B\cos C}} - \dfrac{{\left( {\cos C - \sin A\sin B} \right)}}{{\cos A\cos B}}\]
Now from equation (1) $C = \pi - \left( {A + B} \right)$ so we have,
\[ \Rightarrow \dfrac{1}{{\cos A\cos B\cos C}} - \dfrac{{\left( {\cos \left( {\pi - \left( {A + B} \right)} \right) - \sin A\sin B} \right)}}{{\cos A\cos B}}\]
Now as we know that $\cos \left( {\pi - \theta } \right) = - \cos \theta $ so we have,
\[ \Rightarrow \dfrac{1}{{\cos A\cos B\cos C}} - \dfrac{{\left( { - \cos \left( {A + B} \right) - \sin A\sin B} \right)}}{{\cos A\cos B}}\]
\[ \Rightarrow \dfrac{1}{{\cos A\cos B\cos C}} + \dfrac{{\cos \left( {A + B} \right) + \sin A\sin B}}{{\cos A\cos B}}\]
Now as we know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ so use this we have,
\[ \Rightarrow \dfrac{1}{{\cos A\cos B\cos C}} + \dfrac{{\cos A\cos B - \sin A\sin B + \sin A\sin B}}{{\cos A\cos B}}\]
\[ \Rightarrow \dfrac{1}{{\cos A\cos B\cos C}} + \dfrac{{\cos A\cos B}}{{\cos A\cos B}}\]
\[ \Rightarrow 1 + \dfrac{1}{{\cos A\cos B\cos C}}\]
Now use $\dfrac{1}{{\cos x}} = \sec x$ so we have,
\[ \Rightarrow 1 + \sec A\sec B\sec C\]
= R.H.S
Hence proved.

Note – It is always advisable to remember the direct trigonometric identities like some of them are being mentioned above. Other important identities include${\sin ^2}x + {\cos ^2}x = 1,{\text{ 1 + ta}}{{\text{n}}^2}x = {\sec ^2}x$, ${\text{tanx = }}\dfrac{{\sin x}}{{\cos x}},\sec x = \dfrac{1}{{\cos x}},\cos ecx = \dfrac{1}{{\sin x}}$. These identities help save a lot of time, for these kinds of problems.