Write the values of x for which 2tan^-1x
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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
I hope it help you....
please mark it as a brainieist answer
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
I hope it help you....
please mark it as a brainieist answer
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