Tan(a - b) is just one of the essential trigonometric identities, additionally known as the tangent subtraction formula, provided in trigonometry to discover the worth of the tangent trigonometric function for the difference of angles. Us can uncover the growth of tan(a - b) to represent the tan the a link angle in terms of tangent trigonometric duty for individual angles. Permit us know the expansion of tan(a-b) identity and also its proof in information in the following sections.

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 1 What is Tan(a - b) identity in Trigonometry? 2 Tan(a - b) compound Angle Formula 3 Proof of 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) identification in Trigonometry?

Tan(a-b) identification is among the trigonometry identities for compound angles. The is used when the angle for which the worth of the tangent duty is to be calculated is given in the form of the difference of any type of two angles. The angle (a-b) in the formula the tan(a-b) represents the compound angle.

## tan(a - b) compound Angle Formula

Tan(a - b) formula because that the link angle (a-b) is described as the tangent individually formula in trigonometry. The tan(a-b) formula have the right to be offered as,

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

## Proof of Tan(a - b) identification 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 also denominator through 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 offer the proof of development of tan(a-b) formula using the geometrical building and construction method. Let us see the stepwise source of the formula because that the tangent trigonometric function of the distinction of 2 angles. In the geometrical evidence of tan(a-b) formula, let us at first assume that 'a', 'b', and (a - b), i.e., (a > b). However this formula, in general, is true for any kind of value that a and also b.

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

Construction: i think a right-angled triangle PRQ with ∠PQR = a and also base QR that unit length, as presented in the figure below. Take it a suggest S on PR, such that ∠SQR = b, forming another right-angled triangle SRQ. Extend QR to allude U and from this point, U, attract a perpendicular UT on PQ.

Proof: using trigonometric formulas 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 appropriate triangle STP, ST = cos a(tan a - tan b)

Evaluating the direct pair formed at allude S and also applying the edge sum residential or commercial property of a triangle, us get, ∠RSU = a.Also, ∠PST = a

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

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

Finally, in appropriate 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 apply the development of tan(a - b) for finding the worth of the tangent trigonometric function for angles that can be stood for as the distinction of standard angles in trigonometry. Allow us have a look at the below-given actions to discover the applications of tan(a - b) identity. Take it the example of tan(60º - 45º) to recognize this better.

Step 1: compare the tan(a - b) expression v the given expression to recognize the angles 'a' and 'b'. Here, a = 60º and b = 45º.Step 2: we 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 have the right to compare this v the worth of tan 15º = (√3 - 1)/(√3 + 1). Therefore the an outcome is verified.

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