If sin-1((2a/1+a2))+cos-1((1-a2/1+a2)) = tan-1 ((2x/1-x2)), where a, x ∈ ]0, 1[, then the value of x is

Question

If sin-1((2a/1+a2))+cos-1((1-a2/1+a2)) = tan-1 ((2x/1-x2)), where a, x ∈ ]0, 1[, then the value of x is (A) 0 (B) (a/2) (C) a (D) (2a/1+a2)

Tardigrade Admin · Accepted Answer

sin-1((2a/1+a2))+cos-1((1-a2/1+a2)) = tan-1 ((2x/1-x2)) ⇒ 2tan-1a + 2tan-1a = 2tan-1x ⇒ 4tan-1(a) = 2tan-1x ⇒ 2tan-1a = tan-1x ⇒ tan-1 ((2a/1+a2))= tan-1x ⇒ x = (2a/1+a2)

Q. If $s i n^{- 1} (\frac{2 a}{1 + a ^{2}}) + co s^{- 1} (\frac{1 - a ^{2}}{1 + a ^{2}}) = t a n^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $a, x \in] 0, 1 [$ , then the value of $x$ is

$0$

$\frac{a}{2}$

$a$

$\frac{2 a}{1 + a ^{2}}$

Q. If sin−1(1+a22a​)+cos−1(1+a21−a2​)=tan−1(1−x22x​), where a,x∈]0,1[, then the value of x is

0

2a​

a

1+a22a​

sin−1(1+a22a​)+cos−1(1+a21−a2​)<br/>=tan−1(1−x22x​) ⇒2tan−1a+2tan−1a=2tan−1x ⇒4tan−1(a)=2tan−1x ⇒2tan−1a=tan−1x ⇒tan−1(1+a22a​)=tan−1x ⇒x=1+a22a​

Q. If $s i n^{- 1} (\frac{2 a}{1 + a ^{2}}) + co s^{- 1} (\frac{1 - a ^{2}}{1 + a ^{2}}) = t a n^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $a, x \in] 0, 1 [$ , then the value of $x$ is

$0$

$\frac{a}{2}$

$a$

$\frac{2 a}{1 + a ^{2}}$