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2 arctan(x) = arctan(2x1x2) = arcsin(2x1+x2) = arccos(1x21+x2)

We will learn how to prove the property of the inverse trigonometric function, 2 arctan(x) = arctan(2x1x2) = arcsin(2x1+x2) = arccos(1x21+x2)

or, 2 tan1 x = tan1 (2x1x2) = sin1 (2x1+x2) = cos1 (1x21+x2)

Proof: 

Let, tan1 x = θ         

Therefore, tan θ = x

We know that,

tan 2θ = 2tanθ1tan2θ

tan 2θ = 2x1x2

2θ = tan1(2x1x2)

2 tan1 x = tan1(2x1x2) …………………….. (i)

Again, sin 2θ = 2tanθ1+tan2θ

sin 2θ = 2x1+x2

2θ = sin1(2x1+x2 )

2 tan1 x = sin1(2x1+x2) …………………….. (ii)

Now,  cos 2θ = 1tan2θ1+tan2θ

 cos 2θ =  1x21+x2

2θ = cos1 (1x21+x2)

2 tan1 x = cos (1x21+x2) …………………….. (iii)

Therefore, from (i), (ii) and (iii) we get, 2 tan1 x = tan1 2x1x2 = sin1 2x1+x2 = cos1 1x21+x2                   Proved.

 

Solved examples on property of inverse circular function 2 arctan(x) = arctan(2x1x2) = arcsin(2x1+x2) = arccos(1x21+x2):

1. Find the value of the inverse function tan(2 tan1 15).

Solution:

tan (2 tan1 15)

= tan (tan1 2×151(15)2), [Since, we know that, 2 tan1 x = tan1(2x1x2)]

 = tan (tan1 251125)

= tan (tan1 512)

= 512

 

2. Prove that, 4 tan1 15 - tan1 170 + tan1 199 = π4

Solution:

L. H. S. = 4 tan1 15 - tan1 170 + tan1 199

= 2(2 tan1 15) - tan1 170 + tan1 199

= 2(tan1 2×151(15)2) - tan1 170 + tan1 199, [Since, 2 tan1 x = tan1(2x1x2)]

= 2 (tan1 2151(125))- tan1 170 + tan1 199,

= 2 tan1 512 - (tan1 170 - tan1 199)

= tan1 (2×5121(512)2) - tan1 (1701991+177×199)

= tan1 120199 - tan1 296931

= tan1 120199 - tan1 1239

= tan1 (12019912391+120119×1239)

= tan1 1

= tan1 (tan π4)

= π4 = R. H. S.              Proved.

 Inverse Trigonometric Functions






11 and 12 Grade Math

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