Question

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Answer 1

$\large\bf\underline{Question:-}$

If ,A ,B C are interior angles of a triangle ABC then Prove that tan (B + C )/2 = Cot (A/2)

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$\large\bf\underline{Given:-}$

• A, B ,C are interior angles of triangle ∆ABC.

$\large\bf\underline {To \: Prove:-}$

• tan (B + C )/2 = Cot (A/2)

$\huge\bf\underline{Solution:-}$

• A ,B and C are interior angles of a triangle ABC.

As we know that,:-

≫ sum of interior angles of a triangle = 180°

So,

⠀⠀⠀⠀⠀➝ A + B + C = 180

⠀⠀⠀⠀⠀➝ B + C = 180 - A

On Divide both side by 2,

we get,

⠀⠀⠀⠀⠀➝ (B + C)/2 = (180-A)/2

⠀⠀⠀⠀⠀➝ tan(B + C)/2 = tan (90 -A)/2

• ≫ tan 90 - ∅ = Cot ∅

⠀⠀⠀⠀⠀➝ tan(B + C)/2 = cot (A/2)

Hence, Proved .

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$\boxed{ \begin{minipage}{5cm}\bf\:Some important identities\\tan(90-\theta) = cot \theta\\\bf\:sin (90 - \theta ) = cos \theta\\\bf\:cosec(90 - \theta ) = sec \theta \end{minipage}}$

Answer 2

Answer:

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