from a handpicked tutor in LIVE 1to1 classes
Tan(a  b)
Tan(a  b) is one of the important trigonometric identities, also known as the tangent subtraction formula, used in trigonometry to find the value of the tangent trigonometric function for the difference of angles. We can find the expansion of tan(a  b) to represent the tan of a compound angle in terms of tangent trigonometric function for individual angles. Let us understand the expansion of tan(ab) identity and its proof in detail in the following sections.
What is Tan(a  b) Identity in Trigonometry?
Tan(ab) identity is one of the trigonometry identities for compound angles. It is applied when the angle for which the value of the tangent function is to be calculated is given in the form of the difference of any two angles. The angle (ab) in the formula of tan(ab) represents the compound angle.
tan(a  b) Compound Angle Formula
Tan(a  b) formula for the compound angle (ab) is referred to as the tangent subtraction formula in trigonometry. The tan(ab) formula can be given as,
tan(a  b) = (tan a  tan b)/(1 + tan a·tan b)
Proof of Tan(a  b) Identity Using Sin (a  b) and Cos (a  b)
We can prove the expansion of tan(a  b) given as, tan(a  b) = (tan a  tan b)/(1 + tan a·tan b) using the expansion of sin (a  b) and 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, we get
tan(a  b) = (tan a  tan b)/(1 + tan a·tan b)
Hence, proved.
Geometrical Proof of Tan(a  b) Formula
We can give the proof of expansion of tan(ab) formula using the geometrical construction method. Let us see the stepwise derivation of the formula for the tangent trigonometric function of the difference of two angles. In the geometrical proof of tan(ab) formula, let us initially assume that 'a', 'b', and (a  b), i.e., (a > b). But this formula, in general, is true for any value of a and b.
To prove: tan (a  b) = (tan a  tan b)/(1 + tan a·tan b)
Construction: Assume a rightangled triangle PRQ with ∠PQR = a and base QR of unit length, as shown in the figure below. Take a point S on PR, such that ∠SQR = b, forming another rightangled triangle SRQ. Extend QR to point U and from this point, U, draw a perpendicular UT on PQ.
Proof: Using trigonometric formulas on the rightangled triangle PRQ we get,
tan a = PR/QR
⇒ PR = QR tan a
⇒ PR = tan a (∵ QR = 1)
In righttriangle SRQ,
tan b = SR/QR
⇒ SR = QR tan b
⇒ SR = tan b
⇒ PS = PR  SR = tan a  tan b
⇒ From right triangle STP, ST = cos a(tan a  tan b)
Evaluating the linear pair formed at point S and applying the angle sum property of a triangle, we get, ∠RSU = a.
Also, ∠PST = a [Vertically opposite angles]
From right triangle URS,
tan a = RU/SR
⇒ RU = tan a tan b
⇒ From right 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 apply the expansion of tan(a  b) for finding the value of the tangent trigonometric function for angles that can be represented as the difference of standard angles in trigonometry. Let us have a look at the belowgiven steps to learn the application of tan(a  b) identity. Take the example of tan(60º  45º) to understand this better.
 Step 1: Compare the tan(a  b) expression with the given expression to identify 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 can compare this with the value of tan 15º = (√3  1)/(√3 + 1). Therefore the result is verified.
☛Related Topics on Tan(ab):
Here are some topics that you might be interested in while reading about tan(a  b).
Let us have a look a few solved examples to understand tan(ab) formula better.
Examples Using Tan(a  b)

Example 1: Find the exact value of tan 75º using the expansion of tan(ab).
Solution:
Since, the values of tan function can be easily calculated for 120º and 45º, we can write 75º as (120º  45º).
⇒tan 75º = tan(120º  45º) = (tan 120º  tan 45º)/(1 + tan 120º·tan 45º)º = [(√3)  1]/[1 + (√3)(1)] = (1  √3)/(1  √3) = (√3 + 1)/(√3  1).

Example 2: Apply the tan(a  b) formula to find the expansion of the double angle formula tan 2θ.
Solution:
We can write tan 2θ = tan(θ  (θ))
Applying tan(a  b) = (tan a  tan b)/(1 + tan a·tan b)
tan 2θ = (tan θ  tan (θ))/(1 + tan θ·tan (θ)) = 2tan θ/(1  tan^{2}θ)
∴tan 2θ = 2tan θ/(1  tan^{2}θ)
FAQs on Tan(a  b)
What is Tan(a  b)?
Tan(a  b) is one of the important trigonometric identities also called tan subtraction formula in trigonometry. Tan(a  b) can be given as, tan (a  b) = (tan a  tan b)/(1 + tan a·tan b), where 'a' and 'b' are angles.
What is the Formula of Tan(a  b)?
The tan(ab) formula is used to express the tan compound angle formula in terms of the tangent of individual angles. Tan(a  b) formula in trigonometry can be given as, (tan a  tan b)/(1 + tan a·tan b).
What is Expansion of Tan(a  b)
The expansion of tan(ab) is given as, tan (a  b) = (tan a  tan b)/(1 + tan a·tan b). Here, a and b are the measures of angles.
How to Prove Tan(a  b) Formula?
The proof of tan(ab) formula can be given using the geometrical construction method. We initially assume that 'a', 'b', and (ab) are positive acute angles, i.e., (a  b) < 90. Click here to understand the stepwise method to derive tan(a  b) formula.
What are the Applications of Tan(a  b) Formula?
Tan(a  b) can be used to find the value of tangent function for angles that can be represented as the difference of standard or simpler angles. Thus, making the deduction easier. It can also be used in finding the expansion of other double and multiple angle formulas.
How to Find the Value of tan 15º Using Tan(a  b) Identity.
The value of tan 15º using (a  b) identity can be calculated by first writing it as tan[(45º  30º)] and then applying tan(ab) identity.
⇒tan[(45º  30º)] = (tan 45º  tan 30º)/(1 + tan 45º·tan30º) = [1  (1/√3)]/[1 + (1)(1/√3)] = [(√3  1)/√3]/[(√3 + 1)/√3] = (√3  1)/(√3 + 1).
visual curriculum