If sin-1((2α/1+α 2))+sin-1((2β /1+β 2)) =2tan-1x, then x =

Question

If sin-1((2α/1+α 2))+sin-1((2β /1+β 2)) =2tan-1x, then x = (A) α/β (B) β/α (C) (α + β /1+ α β ) (D) (α + β /1- α β )

Tardigrade Admin · Accepted Answer

sin-1((2α/1+α 2))+sin-1((2β /1+β 2)) =2tan-1x ⇒ 2tan-1α + 2tan-1 β = 2tan-1x ⇒ tan-1α + tan-1β = tan-1x ⇒ tan-1 ((α + β/1- αβ)) = tan-1 x ⇒ x = (α + β /1- α β )

Q. If $s i n^{- 1} (\frac{2 α}{1 + α ^{2}}) + s i n^{- 1} (\frac{2 β}{1 + β ^{2}}) = 2 t a n^{- 1} x$ , then $x =$

$α / β$

$β / α$

$\frac{α + β}{1 + α β}$

$\frac{α + β}{1 - α β}$

Q. If sin−1(1+α22α​)+sin−1(1+β22β​)=2tan−1x, then x=

α/β

β/α

1+αβα+β​

1−αβα+β​

sin−1(1+α22α​)+sin−1(1+β22β​)=2tan−1x ⇒2tan−1α+2tan−1β=2tan−1x ⇒tan−1α+tan−1β=tan−1x ⇒tan−1(1−αβα+β​)=tan−1x ⇒x=1−αβα+β​

Q. If $s i n^{- 1} (\frac{2 α}{1 + α ^{2}}) + s i n^{- 1} (\frac{2 β}{1 + β ^{2}}) = 2 t a n^{- 1} x$ , then $x =$

$α / β$

$β / α$

$\frac{α + β}{1 + α β}$

$\frac{α + β}{1 - α β}$