Question

Prove that cotA+cotB / tanA + tan B + cotB + cotC / tan B +tanC + cot C cotA / tanC +tanA =1

Answer 1

$\dfrac{\cot A+\cot B}{\tan A+\tan B}+\dfrac{\cot B+\cot C}{\tan B+\tan C}+\dfrac{\cot C+\cot A}{\tan C+\tan C}=1$, proved.

Step-by-step explanation:

Prove that, $\dfrac{\cot A+\cot B}{\tan A+\tan B}+\dfrac{\cot B+\cot C}{\tan B+\tan C}+\dfrac{\cot C+\cot A}{\tan C+\tan C}=1$

L.H.S.$=\dfrac{\cot A+\cot B}{\tan A+\tan B}+\dfrac{\cot B+\cot C}{\tan B+\tan C}+\dfrac{\cot C+\cot A}{\tan C+\tan C}$

$=\dfrac{\cot A+\cot B}{\dfrac{1}{\cot A} +\dfrac{1}{\cot B}}+\dfrac{\cot B+\cot C}{\dfrac{1}{\cot B}+\dfrac{1}{\cot C}}+\dfrac{\cot C+\cot A}{\dfrac{1}{\cot C}+\dfrac{1}{\cot A}}$

$={\cot A}{\cot B}+{\cot B}{\cot C}+{\cot C}{\cot A}$     ....(1)

∵ A + B + C = π

⇒ C = π -(A + B)

Now, equation (1) becomes,

$={\cot A}{\cot B}+{\cot B}{\cot (\pi -(A + B)}+{\cot (\pi -(A + B)}{\cot A}$

$={\cot A}{\cot B}-{\cot B}{\cot (A + B)}-{\cot (A + B)}{\cot A}$

$={\cot A}{\cot B}-{\cot (A + B)}[{\cot B}+{\cot A}]$

$={\cot A}{\cot B}-\dfrac{{\cot A}{\cot B}-1}{{\cot A}+{\cot B}} [{\cot B}+{\cot A}]$

$={\cot A}{\cot B}-{\cot A}{\cot B}+1$

= 1 = R.H.S., proved.