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We will learn how to prove the property of the inverse trigonometric function, 2 arctan(x) = arctan(2x1−x2) = arcsin(2x1+x2) = arccos(1−x21+x2)
or, 2 tan−1 x = tan−1 (2x1−x2) = sin−1 (2x1+x2) = cos−1 (1−x21+x2)
Proof:
Let, tan−1 x = θ
Therefore, tan θ = x
We know that,
tan 2θ = 2tanθ1−tan2θ
tan 2θ = 2x1−x2
2θ
= tan−1(2x1−x2)
2 tan−1 x = tan−1(2x1−x2) …………………….. (i)
Again, sin 2θ = 2tanθ1+tan2θ
sin 2θ = 2x1+x2
2θ = sin−1(2x1+x2 )
2 tan−1 x = sin−1(2x1+x2) …………………….. (ii)
Now, cos 2θ = 1−tan2θ1+tan2θ
cos 2θ = 1−x21+x2
2θ = cos−1 (1−x21+x2)
2 tan−1 x = cos (1−x21+x2) …………………….. (iii)
Therefore, from (i), (ii) and (iii) we get, 2 tan−1 x = tan−1 2x1−x2 = sin−1 2x1+x2 = cos−1 1−x21+x2 Proved.
Solved examples on property of inverse circular function 2 arctan(x) = arctan(2x1−x2) = arcsin(2x1+x2) = arccos(1−x21+x2):
1. Find the value of the inverse function tan(2 tan−1 15).
Solution:
tan (2 tan−1 15)
= tan (tan−1 2×151−(15)2), [Since, we know that, 2 tan−1 x = tan−1(2x1−x2)]
= tan (tan−1 251−125)
= tan (tan−1 512)
= 512
2. Prove that, 4 tan−1 15 - tan−1 170 + tan−1 199 = π4
Solution:
L. H. S. = 4 tan−1 15 - tan−1 170 + tan−1 199
= 2(2 tan−1 15) - tan−1 170 + tan−1 199
= 2(tan−1 2×151−(15)2) - tan−1 170 + tan−1 199, [Since, 2 tan−1 x = tan−1(2x1−x2)]
= 2 (tan−1 2151−(125))- tan−1 170 + tan−1 199,
= 2 tan−1 512 - (tan−1 170 - tan−1 199)
= tan−1 (2×5121−(512)2) - tan−1 (170−1991+177×199)
= tan−1 120199 - tan−1 296931
= tan−1 120199 - tan−1 1239
= tan−1 (120199−12391+120119×1239)
= tan−1 1
= tan−1 (tan π4)
= π4 = R. H. S. Proved.
● Inverse Trigonometric Functions
11 and 12 Grade Math
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