Chapter 8: Q. 61 (page 531)

Law of Tangents For any triangle, derive the Law of Tangents.
$\frac{a - b}{a + b} = \frac{\tan [\frac{1}{2} (A - B)]}{\tan [\frac{1}{2} (A + B)]}$

Short Answer

Expert verified

It is proved that $\frac{a - b}{a + b} = \frac{\tan [\frac{1}{2} (A - B)]}{\tan [\frac{1}{2} (A + B)]}$

Step by step solution

Step 1. Given Information

Mollweide's formula $\frac{a + b}{c} = \frac{\cos [\frac{1}{2} (A - B)]}{\sin (\frac{1}{2} C)} - - - - - (1) \frac{a - b}{c} = \frac{\sin [\frac{1}{2} (A - B)]}{\cos (\frac{1}{2} C)} - - - - - - (2)$

Sum of angles in a triangle $A + B + c = 180^{\circ}$

Step 2. Multiply equation (2) with the reciprocal of equation (1)

$\frac{a - b}{c} \times \frac{c}{a + b} = \frac{\sin [\frac{1}{2} (A - B)]}{\cos (\frac{1}{2} C)} \times \frac{\sin (\frac{1}{2} C)}{\cos [\frac{1}{2} (A - B)]} \frac{a - b}{a + b} = \frac{\sin [\frac{1}{2} (A - B)]}{\cos [\frac{1}{2} (A - B)]} \times \frac{\sin (\frac{1}{2} C)}{\cos (\frac{1}{2} C)}$

Step 3. Finding C2

$A + B + C = 180 C = 180 - A - B \frac{C}{2} = 90 - (\frac{A + B}{2})$

Step 4. Substitute C2=90∘-(A+B)2 into sinC2

$\sin \frac{C}{2} = \sin (90 - \frac{A + B}{2}) = \cos (\frac{A + B}{2})$

Step 5. Substitute C2=90∘-(A+B)2into cosC2.

Step 6. Plug sinA+B2 & cosA+B2into a-ba+b

$\frac{a - b}{a + b} = \frac{\sin \frac{A - B}{2}}{\cos \frac{A - B}{2}} \times \frac{\cos \frac{A + B}{2}}{\sin \frac{A + B}{2}} \frac{a - b}{a + b} = \tan (\frac{A - B}{2}) \times \frac{1}{\tan (\frac{A + B}{2})} \frac{a - b}{a + b} = \frac{\tan (\frac{A - B}{2})}{\tan (\frac{A + B}{2})}$

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Law of Tangents For any triangle, derive the Law of Tangents.
$\frac{a - b}{a + b} = \frac{\tan [\frac{1}{2} (A - B)]}{\tan [\frac{1}{2} (A + B)]}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Multiply equation (2) with the reciprocal of equation (1)

Step 3. Finding C2

Step 4. Substitute C2=90∘-(A+B)2 into sinC2

Step 5. Substitute C2=90∘-(A+B)2into cosC2.

Step 6. Plug sinA+B2 & cosA+B2into a-ba+b

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Law of Tangents For any triangle, derive the Law of Tangents.a-ba+b=tan12A-Btan12A+B

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Multiply equation (2) with the reciprocal of equation (1)

Step 3. Finding C2

Step 4. Substitute C2=90∘-(A+B)2 into sinC2

Step 5. Substitute C2=90∘-(A+B)2into cosC2.

Step 6. Plug sinA+B2 & cosA+B2into a-ba+b

One App. One Place for Learning.

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Law of Tangents For any triangle, derive the Law of Tangents.
$\frac{a - b}{a + b} = \frac{\tan [\frac{1}{2} (A - B)]}{\tan [\frac{1}{2} (A + B)]}$