Tan(a - b) is one of the important trigonometric identities, also known as the tangent individually formula, provided in trigonometry to find the value of the tangent trigonometric duty for the difference of angles. We can discover the development of tan(a - b) to represent the tan of a link angle in regards to tangent trigonometric duty for individual angles. Permit us recognize the growth of tan(a-b) identity and its proof in detail in the following sections.

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1.What is Tan(a - b) identification in Trigonometry?
2.Tan(a - b) compound Angle Formula
3.Proof that Tan(a - b) Identity
4.Geometrical evidence of Tan(a - b) Formula
5.How to apply Tan(a - b)?
6.FAQs on Tan(a - b)

What is Tan(a - b) identity in Trigonometry?


Tan(a-b) identity is among the trigonometry identities for link angles. It is used when the angle because that which the value of the tangent duty is to be calculated is given in the kind of the distinction of any type of two angles. The edge (a-b) in the formula that tan(a-b) to represent the compound angle.


tan(a - b) link Angle Formula


Tan(a - b) formula because that the link angle (a-b) is referred to as the tangent subtraction formula in trigonometry. The tan(a-b) formula deserve to be offered as,

tan(a - b) = (tan a - tan b)/(1 + tan a·tan b)

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Proof the Tan(a - b) identity Using Sin (a - b) and also Cos (a - b)


We can prove the expansion of tan(a - b) provided as, tan(a - b) = (tan a - tan b)/(1 + tan a·tan b) making use of the development of sin (a - b) and also cos (a - b).we know, tan(a - b) = sin(a - b)/cos(a - b)

= (sin a cos b - cos a sin b)/(cos a cos b + sin a sin b)

Dividing the numerator and denominator by cos a cos b, us get

tan(a - b) = (tan a - tan b)/(1 + tan a·tan b)

Hence, proved.


Geometrical proof of Tan(a - b) Formula


We can provide the proof of expansion of tan(a-b) formula making use of the geometrical building and construction method. Let us see the stepwise source of the formula for the tangent trigonometric role of the distinction of 2 angles. In the geometrical proof of tan(a-b) formula, allow us at first assume that 'a', 'b', and also (a - b), i.e., (a > b). But this formula, in general, is true for any value the a and also b.

To prove: tan (a - b) = (tan a - tan b)/(1 + tan a·tan b)

Construction: assume a right-angled triangle PRQ through ∠PQR = a and base QR the unit length, as displayed in the number below. Take a allude S on PR, such the ∠SQR = b, forming an additional right-angled triangle SRQ. Extend QR to allude U and from this point, U, draw a perpendicular UT top top PQ.

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Proof: using trigonometric recipe on the right-angled triangle PRQ us get,tan a = PR/QR⇒ PR = QR tan a⇒ PR = tan a (∵ QR = 1)

In right-triangle SRQ,tan b = SR/QR⇒ SR = QR tan b⇒ SR = tan b

⇒ PS = PR - SR = tan a - tan b

⇒ From best triangle STP, ST = cos a(tan a - tan b)

Evaluating the linear pair developed at point S and also applying the edge sum residential or commercial property of a triangle, we get, ∠RSU = a.Also, ∠PST = a

From ideal triangle URS,tan a = RU/SR⇒ RU = tan a tan b

⇒ From ideal triangle UTQ, QT = cos a(QU) = cos a(QR + RU) = cos a(1 + tan a tan b)

Finally, in right triangle STQ,

tan(a - b) = ST/TQ = cos a(tan a - tan b)/cos a(1 + tan a tan b) = (tan a - tan b)(1 + tan a tan b)

Hence, proved.


How to apply Tan(a - b)?


We can use the expansion of tan(a - b) for finding the worth of the tangent trigonometric role for angles that can be represented as the difference of typical angles in trigonometry. Let us have a look in ~ the below-given procedures to learn the applications of tan(a - b) identity. Take the example of tan(60º - 45º) to know this better.

Step 1: compare the tan(a - b) expression with the provided expression to identify the angle 'a' and also 'b'. Here, a = 60º and also b = 45º.Step 2: us know, tan(a - b) = (tan a - tan b)/(1 + tan a·tan b)⇒ tan(60º - 45º) = (tan 60º - tan 45º)/(1 + tan 60º·tan 45º)since, tan 60º = √3, tan 45º = 1⇒ tan(60º - 45º) = <√3 - 1>/<1 + (√3)·1> = (√3 - 1)/(√3 + 1).Also, we deserve to compare this through the worth of tan 15º = (√3 - 1)/(√3 + 1). As such the an outcome is verified.

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