TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 28 of 160

Marks: +1, -0

In a ΔABC,

\cot A+\cot B+\cot C=

Solution:

Let a triangle

A B C

of sides

a, b, c

, having area

\Delta

Area

=\Delta

\frac{1}{2} b c \sin A = \Delta

....(i)

and

\frac{1}{2} a c \sin B = \Delta

....(ii)

and

\frac{1}{2} a b \sin c = \Delta

....(iii)

Using cosine rule

a^{2}=b^{2}+c^{2}-2 b c \cos A

and

b^{2}=a^{2}+c^{2}-2 a c \cos B

and

c^{2}=a^{2}+b^{2}-2 a b \cos C

On adding, we get

a^{2}+b^{2}+c^{2}=2 a^{2}+2 b^{2}+2 c^{2}-2 a b \cos C-2 a c \cos B-2 b c \cos A

a^{2}+b^{2}+c^{2}=2(a b \cos C + a c \cos B + b c \cos A)

...(iv)

Now, from

Eq. (i)

bc = \frac{2 \Delta}{\sin A}

Eq. (ii),

ac = \frac{2 \Delta}{\sin B}

Eq. (iii),

ab = \frac{2 \Delta}{\sin C}

Putting these values in Eq. (iv), we get

a^{2}+b^{2}+c^{2}=2\left( \frac{2 \Delta}{\sin C} \cos C + \frac{2 \Delta}{\sin B} \cos B + \frac{2 \Delta}{\sin A} \cos A \right)

a^{2}+b^{2}+c^{2}=4 \Delta (\cot C + \cot B + \cot A)

\cot C + \cot B + \cot A = \frac{a^{2}+b^{2}+c^{2}}{4 \Delta}

\cot A + \cot B + \cot C = \frac{a^{2}+b^{2}+c^{2}}{4 \Delta}

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