sin[ tan-1(1-x2/2x)+ cos-1(1-x2/1+x2)] is equal to

Question

sin[ tan-1(1-x2/2x)+ cos-1(1-x2/1+x2)] is equal to (A) 0 (B) 1 (C) (1/√2) (D) √2

Tardigrade Admin · Accepted Answer

Put x= tan θ ∴ sin [tan-1 (1-x2/2x) + cos-1 (1-x2/1+x2)] =sin[tan-1(cot 2 θ)+ cos-1 cos 2 θ] = sin [tan-1 tan((π/2) -2 θ)+2 θ] = sin((π/2)-2 θ+2 θ) = sin (π/2) = 1

Q. $sin [tan^{- 1} \frac{1 - x ^{2}}{2 x} + cos^{- 1} \frac{1 - x ^{2}}{1 + x ^{2}}]$ is equal to

0

1

$\frac{1}{2}$

$2$

Put $x = t an θ$
$∴ s in [t a n^{- 1} \frac{1 - x ^{2}}{2 x} + co s^{- 1} \frac{1 - x ^{2}}{1 + x ^{2}}]$
$= s in [t a n^{- 1} (co t 2 θ) + co s^{- 1} cos 2 θ]$
$= s in [t a n^{- 1} t an (\frac{π}{2} - 2 θ) + 2 θ]$
$= s in (\frac{π}{2} - 2 θ + 2 θ)$
$= s in \frac{π}{2} = 1$

Q. sin[tan−12x1−x2​+cos−11+x21−x2​] is equal to

0

1

2​1​

2​

Put x=tanθ ∴sin[tan−12x1−x2​+cos−11+x21−x2​] =sin[tan−1(cot2θ)+cos−1cos2θ] =sin[tan−1tan(2π​−2θ)+2θ] =sin(2π​−2θ+2θ) =sin2π​=1

Q. $sin [tan^{- 1} \frac{1 - x ^{2}}{2 x} + cos^{- 1} \frac{1 - x ^{2}}{1 + x ^{2}}]$ is equal to

$\frac{1}{2}$

$2$

Put $x = t an θ$
$∴ s in [t a n^{- 1} \frac{1 - x ^{2}}{2 x} + co s^{- 1} \frac{1 - x ^{2}}{1 + x ^{2}}]$
$= s in [t a n^{- 1} (co t 2 θ) + co s^{- 1} cos 2 θ]$
$= s in [t a n^{- 1} t an (\frac{π}{2} - 2 θ) + 2 θ]$
$= s in (\frac{π}{2} - 2 θ + 2 θ)$
$= s in \frac{π}{2} = 1$