Question

In a ∆ABC,
$\tt b^2 + c^2 = 1999a^2$
then, find the value of
$\tt \dfrac{\cot B + \cot C}{\cot A}$
​

Answer 1

Step-by-step explanation:

Given :-

In ∆ ABC , b²+c² = 1999a²

To find :-

The value of (cot B + cot C)/cot A

Solution :-

Given that

In ∆ ABC , b²+c² = 1999a²----------(1)

We know that

Area of a triangle by Heron's formula

∆ = √[S(S-a)(S-b)(S-c)] sq.units -----(2)

Now,

cot A = cos A/sin A

We know that

cos A = (b²+c²-a²)/2bc and sin A = 2∆/bc

Therefore,

cot A = [(b²+c²-a²)/2bc]/(2∆/bc )

On cancelling bc in both numerator and denominator then

cot A = [(b²+c²-a²)/2]/2∆

=> cot A = (b²+c²-a²)/(2×2∆)

=> cot A = (b²+c²-a²)/4∆--------------(3)

and

cot B = cos B / sin B

We know that

cos B = (c²+a²-b²)/2ca and sin B = 2∆/ca

Therefore,

cot B = [(c²+a²-b²)/2ca] / (2∆/ca)

On cancelling ca in both numerator and denominator then

=> cot B = [(c²+a²-b²)/2]/2∆

=> cot B = (c²+a²-b²)/(2×2∆)

=> cot B = (c²+a²-b²)/4∆ -------------(4)

and

cot C = cos C / sin C

we know that

cos C =( a²+b²-c²)/2ab and sin C = 2∆/ab

Now,

Therefore,

cot C = [ (a²+b²-c²)/2ab]/(2∆/ab)

On cancelling ab in both numerator and denominator then

=> cot C = [(a²+b²-c²)/2]/2∆

=> cot C = (a²+b²-c²)/(2×2∆)

=> cot C = (a²+b²-c²)/4∆ -------------(5)

On adding (4) and (5) then

cot B + cot C

= [(c²+a²-b²)/4∆]+[(a²+b²-c²)/4∆]

=> cot B + cot C = (c²+a²-b²+a²+b²+c²)/4∆

=> cot B + cot C = (c²-c²+b²-b²+a²+a²)/4∆

=> cot B + cot C = (0+0+2a²)/4∆

=> cot B + cot C = 2a²/4∆ -------------(6)

On dividing (6) by (3) then

(cot B + cot C)/cot A

=> (2a²/4∆)/[(b²+c²-a²)/4∆]

=> (2a²/4∆) ×[4∆/(b²+c²-a²)]

On cancelling 4∆ then

=> 2a²/(b²+c²-a²)

On dividing both numerator and denominator with a² then

=> (2a²/a²)/[(b²+c²-a²)/a²]

=> 2/[{(b²+c²)/a²}-(a²/a²)]

=> 2/[{(b²+c²)/a²}-1]

From (1) we have

=> 2/[(1999a²/a²)-1]

=> 2/(1999-1)

=> 2/1998

=> 1/999

Therefore, (cot B + cot C)/cot A = 1/999

Answer :-

The value of (cot B + cot C)/cot A is 1/999

Used formulae:-

→ Area of a triangle by Heron's formula

∆ = √[S(S-a)(S-b)(S-c)] sq.units

• Where, S = (a+b+c)/2 units
• a,b and c are the sides

→ cos A = (b²+c²-a²)/2bc

→ cos B = (c²+a²-b²)/2ca

→ cos C = (a²+b²-c²)/2ab

→ sin A = 2∆/bc

→ sin B = 2∆/ca

→ sin C = 2∆/ab

→ Cot θ = Cos θ / Sin θ

Answer 2

Given:-

• b² + c² = 1999a².

To Find:-

$\sf \large The\: value \: of \: \frac{cotB + cotC}{cotA} .$

Solution:-

$\sf \large cotA = \frac{cosA}{sinA} = \frac{cosA}{ \frac{2 ∆}{bc} }$

$\sf \large = \frac{bc \: \: cosA}{2 ∆}$

$\sf \large = \frac{bc \bigg [ \frac{b {}^{2} + c {}^{2} - a {}^{2} }{2bc} \bigg ]}{2 ∆}$

$\sf \large = \frac{b {}^{2} + c {}^{2} - a {}^{2} }{4 ∆}$

$\sf \large = cotB = \frac{cosB}{sinB} = \frac{cosB}{ \frac{2 ∆}{ac}} = \frac{ac \: \: cosB}{2 ∆}$

$\sf \large = \frac{c {}^{2} + a {}^{2} - b {}^{2} }{4 ∆}$

$\sf \large = cotC = \frac{cosC}{sinC} = \frac{cosC}{ \frac{2 ∆}{ab} }$

$\sf \large = \frac{ab \: \: cos C}{2 ∆}$

$\sf \large = \frac{a {}^{2} + b {}^{2} - c {}^{2} }{4 ∆}$

substituting the above in

$\sf \large \frac{cotB + cotC }{cotA}$we have,

$\sf \large \frac{cotB + cotC}{cotA} = \frac{ \frac{c {}^{2} + a {}^{2} - b {}^{2} }{4 ∆} + \frac{a {}^{2} + b {}^{2} - c {}^{2} }{4 ∆} }{ \frac{b {}^{2} + c {}^{2} - a {}^{2} }{4 ∆} }$

On simplification, we get

$\sf \large = \frac{2a {}^{2} }{b {}^{2} + c {}^{2} - a {}^{2} }$

$\sf \large = \frac{2}{ \frac{b {}^{2} + c {}^{2} }{a {}^{2} } - 1 }$

$\sf \large = \frac{2}{1999 - 1} = \frac{2}{1998} = \frac{1}{999}$

Answer:-

$\sf \large \red{\frac{cotB + cotC}{cotA} = \frac{1}{999} .}$

Hope you have satisfied. ⚘