Prove that 1-cos A-cos B-cos A-B - 1-cos A-cos B-cos A-B-tan A - 2 B - 2

take the numerator of L.H.S 1 cosA+cosB cos(A+B) simplifying 1 cosA = 2sin2 A/2 by half angle identity cosB cos(A+B) = 2sin(A/2)sin((A+2B)/2) differe ...

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Byju's Answer

Standard XII

Mathematics

Linear Functions

Prove that 1-...

Question

Prove that $\frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)} = tan \frac{A}{2} cot (\frac{B}{2})$ .

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Solution

Consider, LHS $= \frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)}$
$= \frac{2 {sin}^{2} (\frac{A}{2}) + 2 sin (\frac{A}{2}) sin (\frac{A}{2} + B)}{2 {cos}^{2} (\frac{A}{2}) - 2 cos (\frac{A}{2}) cos (\frac{A}{2} + B)}$ [Since, $cos C - cos D = 2 sin \frac{(D - C)}{2} sin \frac{(D + C)}{2}$ ]
$= \frac{2 sin (\frac{A}{2})}{2 cos (\frac{A}{2})} ⎛ ⎝ \frac{sin (\frac{A}{2}) + sin (\frac{A}{2} + B)}{cos (\frac{A}{2}) - cos (\frac{A}{2} + B)} ⎞ ⎠$
$= tan (\frac{A}{2}) ⎛ ⎝ \frac{2 sin (\frac{A + B}{2}) cos (\frac{B}{2})}{2 sin (\frac{A + B}{2}) sin (\frac{B}{2})} ⎞ ⎠$
[Since, $sin C + sin D = 2 sin \frac{(C + D)}{2} sin \frac{(C - D)}{2}$ ]
$= tan (\frac{A}{2}) cot (\frac{B}{2})$
$=$ RHS
Therefore, LHS $=$ RHS
Hence, $\frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)} = tan \frac{A}{2} cot (\frac{B}{2})$

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\frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)} =

If tanA/2=±√1-e/1+e*tanB/2

So prove

CosB = cosA-e/1-ecosA

Q. In any triangle ABC,

\frac{1 - cos A + cos B + cos C}{1 - cos C + cos A + cos B} = \frac{tan (A / 2)}{tan (C / 2)}

Q. If

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and

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Prove that 1−cosA+cosB−cos(A+B)1+cosA−cosB−cos(A+B)=tanA2cot(B2).

Prove that $\frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)} = tan \frac{A}{2} cot (\frac{B}{2})$ .