Question

tan A/2=(√1-cosA/1+cosA)​ proof

Answer 1

We have to prove that:

$\tan \dfrac{A}{2} =\sqrt{\dfrac{1-\cos A}{1+\cos A}} .$

R.H.S. = $\sqrt{\dfrac{1-\cos A}{1+\cos A}}$

Using the trigonometric identity:

$1-\cos A = 2\sin^2 \dfrac{A}{2} }$ and

$1+\cos A = 2\cos^2 \dfrac{A}{2} }$

= $\sqrt{\dfrac{2\sin^2 \dfrac{A}{2} }{2\cos^2 \dfrac{A}{2} }}$

= $\sqrt{\dfrac{\sin^2 \dfrac{A}{2} }{\cos^2 \dfrac{A}{2} }}$

= $\sqrt{\tan^2 \dfrac{A}{2}}$

= $\tan \dfrac{A}{2}$

= L.H.S., proved.

∴ $\tan \dfrac{A}{2} =\sqrt{\dfrac{1-\cos A}{1+\cos A}}$, proved.

Answer 2

To Prove :- tan A/2 = √(1 - cosA/1 + cosA) ?

Solution :-

solving RHS,

→ √(1 - cosA/1 + cosA)

→ √{(1 - cosA/1 + cosA) * (1 - cosA/1 - cosA)}

→ √{(1 - cosA)² / (1 - cos²A)}

→ √{(1 - cosA)² / (sin²A)}

→ (1 - cosA) / sinA

putting

• (1 - cosA) = 2sin²A/2
• sinA = 2 * sinA/2 * cosA/2

→ (2sin²A/2) / (2 * sinA/2 * cosA/2)

→ (sinA/2) / (cosA/2)

→ (tanA/2) = LHS.

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