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We will learn how to prove the property of the inverse trigonometric function arctan(x) + arctan(y) = arctan(x+y1βxy), (i.e., tanβ1 x + tanβ1 y = tanβ1 (x+y1βxy) if x > 0, y > 0 and xy < 1.
1. Prove that arctan(x) + arctan(y) = arctan(x+y1βxy), if x > 0, y > 0 and xy < 1.
Proof:
Let, tanβ1 x = Ξ± and tanβ1 y = Ξ²
From tanβ1 x = Ξ± we get,
x = tan Ξ±
and from tanβ1 y = Ξ² we get,
y = tan Ξ²
Now, tan (Ξ± + Ξ²) = (tanΞ±+tanΞ²1βtanΞ±tanΞ²)
tan (Ξ± + Ξ²) = x+y1βxy
β Ξ± + Ξ² = tanβ1 (x+y1βxy)
β tanβ1 x + tanβ1 y = tanβ1 (x+y1βxy)
Therefore, tanβ1 x + tanβ1 y = tanβ1 (x+y1βxy), if x > 0, y > 0 and xy < 1.
2. Prove that arctan(x) + arctan(y) = Ο + arctan(x+y1βxy), if x > 0, y > 0 and xy > 1. And
arctan(x) + arctan(y) = arctan(x+y1βxy) - Ο, if x < 0, y < 0 and xy > 1.
Proof: If x > 0, y > 0 such that xy > 1, then x+y1βxy is positive and therefore, x+y1βxy is positive angle between 0Β° and 90Β°.
Similarly, if x < 0, y < 0 such that xy > 1, then x+y1βxy is positive and therefore, tanβ1 (x+y1βxy) is a negative angle while tanβ1 x + tanβ1 y is a positive angle while tanβ1 x + tanβ1 y is a non-negative angle. Therefore, tanβ1 x + tanβ1 y = Ο + tanβ1 (x+y1βxy), if x > 0, y > 0 and xy > 1 and
arctan(x) + arctan(y) = arctan(x+y1βxy) - Ο, if x < 0, y < 0 and xy > 1.
Solved examples on property of inverse circular function tanβ1 x + tanβ1 y = tanβ1 (x+y1βxy)
1. Prove that 4 (2 tanβ1 13 + tanβ1 17) = Ο
Solution:
2 tanβ1 13
= tanβ1 13 + tanβ1 13
= tanβ1 (13+131β13β’13)
= tanβ1 34
Now L. H. S. = 4 (2 tanβ1 13 + tanβ1 17)
= 4 (tanβ1 34 + tanβ1 17)
= 4 tanβ1 (34+171β34β’17)
= 4 tanβ1 (2528 x 2825)
= 4 tanβ1 1
= 4 Β· Ο4
= Ο = R.H.S. Proved.
2. Prove that, tanβ1 14 + tanβ1 29 + tanβ1 15 + tanβ1 18 = Ο/4.
Solution:
L. H. S. = tanβ1 14 + tanβ1 29 + tanβ1 15 + tanβ1 18
= tanβ1 14+291β14β’29 + tanβ1 15+181β15β’18
= tanβ1 (1736 x 3634) + tanβ1 (1340 x 4039)
= tanβ1 12 + tanβ1 13
= tanβ1 12+131β12β’13
= tanβ1 1
= Ο4 = R. H. S. Proved.
β Inverse Trigonometric Functions
11 and 12 Grade Math
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