Law of tangents is a law in trigonometry which relates the sides and angles of a right triangle. Tangent rule gives the relationship between the sum and differences of the sides and angles of a triangle. The tangent rule can be used to find the remaining parts of any triangle for which two sides and one angle or one side and two angles are given. Law of tangents finds extensive applications in various Mathematical computations just as sine and cosine laws. The law of tangents for a triangle with angles A, B and C opposite to the sides a, b and c respectively is given as:
\[\frac{a-b}{a+b}\] = \[\frac{tan(\frac{A-B}{2})}{tan(\frac{A+B}{2})}\] |
The rule of tangent establishes a relationship between the sum and differences of any two sides of a triangle and their corresponding angles. The tangent rule states that the ratio of difference and sum of any two sides of a triangle is equal to the ratio of the tangent of half the difference and tangent of sum of the angles opposite to these sides. Rule of tangents can be used to find the unknown parts of a triangle when two sides and an angle or two angles and a side are given.
The rule of tangents can be proved using the sine rule. Sine rule states that the ratio of any side of a triangle and the sine of the angle opposite to it is a constant. This basic rule is the foundation for proving the rule of tangents.
The ratio of sum and difference of any two sides of a triangle is equal to the ratio of tangent of half the sum and tangent of half the difference of the angles opposite to the corresponding sides.
Data:
In the triangle ABC, ∠A,∠B and ∠C are the angles opposite to the sides ‘a’, ‘b’ and ‘c’ respectively.
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To Prove: What is tangent law?
i.e. \[\frac{a-b}{a+b}\] = \[\frac{tan(\frac{A-B}{2})}{tan(\frac{A+B}{2})}\]
Statement | Reason |
\[\frac{a}{SinA}\] = \[\frac{b}{SinB}\] = \[\frac{c}{SinC}\] | Applying sine rule to the triangle ABC |
\[\frac{a}{SinA}\] = \[\frac{b}{SinB}\] = d | Equating the ratio to a constant |
\[\frac{a}{SinA}\] = d and \[\frac{b}{SinB}\] = d | Equating each ratio to the constant ‘k’ |
a = d Sin A and b = d Sin B | Cross multiplication |
a + b = d Sin A + d Sin B = d (Sin A + Sin B) → (1) | Sum of ‘a’ and ‘b’ |
a - b = d Sin A - d Sin B = d (Sin A - Sin B) → (2) | Difference between ‘a’ and ‘b’ |
Sin M + Sin N = 2 Sin\[(\frac{M+N}{2})\] Cos\[(\frac{M-N}{2})\] Sin M - Sin N = 2 Cos \[(\frac{M+N}{2})\] Sin \[(\frac{M-N}{2})\] | Trigonometric identities |
\[\frac{a-b}{a+b}\] = \[\frac{d(SinA-SinB)}{d(SinA+SinB)}\] = \[\frac{SinA-SinB}{SinA+SinB}\] | Dividing (1) and (2) |
\[\frac{2Cos(\frac{A+B}{2})Sin(\frac{A-B}{2})}{2Sin(\frac{A+B}{2})Cos(\frac{A-B}{2})}\] | Substituting the trigonometric identities in the above equation |
\[\frac{a-b}{a+b}\] = \[\frac{tan(\frac{A-B}{2})}{tan(\frac{A+B}{2})}\] | By definition, Tan R = Sin R / Cos R |
The final equation gives the law of tangent formula.
Consider a triangle with sides ‘f’, ‘g’ and ‘h’ opposite to the vertices F, G and H. The sum of two sides is (f + g) or (g + h) or (h + f). Similarly the difference between two sides is given as (f - g) or (g - h) or (h - f).
The law of tangent formula for the ratio of difference and sum of two sides of a triangle is given as:
\[\frac{F-g}{f+g}\] = \[\frac{tan(\frac{F-G}{2})}{tan(\frac{F+G}{2})}\] → (1)
\[\frac{g-h}{g+h}\] = \[\frac{tan(\frac{G-H}{2})}{tan(\frac{G+H}{2})}\] → (2)
\[\frac{h-f}{h+f}\] = \[\frac{tan(\frac{H-F}{2})}{tan(\frac{H+F}{2})}\] → (3)
Law of tangents for triangles was given by a Persian Mathematician Nasir al-Din al-Tusi in the 13th century. He explained what is law of tangents for spherical triangles.
The spherical law of tangents states that the ratio of tangent of the difference between two sides and the tangent of its sum is equal to the ratio of tangent of the half of the difference between their opposite angles and the tangent of half of their sum. For a triangle with angles P, Q and R opposite to the sides ‘p’, ‘q’ and ‘r’ respectively, the spherical law of tangents is given as:
\[\frac{tan(\frac{a-b}{2})}{tan(\frac{a+b}{2})}\] = \[\frac{tan(\frac{A-B}{2})}{tan(\frac{A+B}{2})}\] |
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