Prove that in ABC- tanB-C-2- cot A-2 -

In a triangle, sum of all the internal angles is 180^∘. A+B+C = 180^∘ A+B+C/2 = 90^∘ B+C/2 = 90^∘ A/2 tan (B+C/2) = tan (90^∘ A/2) Since tan (90^∘ θ) = cot θ ...

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Standard IX

Mathematics

Angle Sum Property of a Triangle

Prove that in...

Question

Prove that in $△ A B C$ , $t a n (\frac{B + C}{2}) = c o t (\frac{A}{2})$ .

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Solution

In a triangle, sum of all the internal angles is 180 $^{\circ}$ .
$A + B + C = 180^{\circ}$
$⟹ \frac{A + B + C}{2} = 90^{\circ}$
$⟹ \frac{B + C}{2} = 90^{\circ} - \frac{A}{2}$
$⟹ t a n (\frac{B + C}{2}) = t a n (90^{\circ} - \frac{A}{2})$

Since
$t a n (90^{\circ} - θ) = c o t θ$

$⟹ t a n (\frac{B + C}{2}) = t a n (90^{\circ} - \frac{A}{2})$

$∵ t a n (90^{\circ} - \frac{A}{2}) = c o t \frac{A}{2}$

$⟹ t a n (\frac{B + C}{2}) = c o t \frac{A}{2}$

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Prove that in △ ABC, tan(B+C2)=cot(A2) .

In a triangle, sum of all the internal angles is 180∘. A+B+C=180∘ ⟹A+B+C2=90∘ ⟹B+C2=90∘−A2 ⟹tan(B+C2)=tan(90∘−A2) Since tan(90∘−θ)=cot θ ⟹tan(B+C2)=tan(90∘−A2) ∵tan(90∘−A2)=cotA2 ⟹tan(B+C2)=cotA2

Prove that in $△ A B C$ , $t a n (\frac{B + C}{2}) = c o t (\frac{A}{2})$ .