Question

Tan inverse alpha tan inverse beta tan inverse gamma

Answer 1

Answer:

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Answer 2

prove the property of the inverse trigonometric function arctan(x) + arctan(y) + arctan(z) = arctanx+y+z−xyz1−xy−yz−zx (i.e., tan−1 x + tan−1 y + tan−1 z = tan−1 x+y+z−xyz1−xy−yz−zx)

Prove that, tan−1 x + tan−1 y + tan−1 z = tan−1 x+y+z–xyz1–xy–yz–zx

Proof :

Let, tan−1 x = α, tan−1 y = β and tan−1γ

Therefore, tan α = x, tan β = y and tan γ = z

We know that, tan (α + β + γ) = tanα+tanβ+tanγ−tanαtanβtanγ1−tanαtanβ−tanβtanγ−tanγtanα

tan (α + β + γ) = x+y+z–xyz1–xy–yz–zx

α + β + γ = tan−1 x+y+z–xyz1–xy–yz–zx

or, tan−1 x + tan−1 y + tan−1 z = tan−1 x+y+z–xyz1–xy–yz–zx. Proved.

Second method:

We can prove tan−1 x + tan−1 y + tan−1 z = tan−1 x+y+z–xyz1–xy–yz–zx in other way.

We know that, tan−1 x + tan−1 y = tan−1 x+y1–xy

Therefore, tan−1 x + tan−1 y + tan−1 z = tan−1 x+y1–xy + tan−1 z

tan−1 x + tan−1 y + tan−1 z = tan−1 x+y1–xy+z1−x+y1−xy∙z

tan−1 x + tan−1 y + tan−1 z = tan−1 x+y+z–xyz1–xy–yz–zx. Proved.