`\color{red} \ox \color{red} (COMMON \ CONFUSION) `
`sin (sin^-1 x) = x, x ∈ [– 1, 1]` and `sin^-1 (sin x) = x, x ∈ [ - pi/2, pi/2]`
`cos (cos^-1 x) = x, x ∈ [– 1, 1]` and `cos ^-1 (cos x) = x, x ∈ [ 0,pi]`
`tan (tan^-1 x) = x, x ∈ R` and `tan^-1 (tan x) = x, x ∈ ( - pi/2, pi/2)`
`cosec (cosec^-1 x) = x, |x| ge 1` and `cosec^-1 (cosec x) = x, x ∈ [ - pi/2, pi/2] , xne0`
`sec (sec^-1 x) = x, |x| ge 1` and `sec^-1 (sec x) = x, x ∈ [ 0, pi], x ne pi/2`
`cot (cot^-1 x) = x, x ∈ R` and `cot^-1 (cot x) = x, x ∈ ( 0,pi)`
`\color{purple}ul(✓✓) \color{purple} "ALERT"`
`sin(sin^(-1)1) = 1`
`sin(sin^(-1)2) ne 2` (Firstly given question is wrong as `2` isn't in domain)
`\color { maroon} ® \color{maroon} ul (" REMEMBER") :` So if `sin(sin^(-1)x)` is given means `x` always will be in domain,So we can write directly equal to `x`
Property 1 :
• `sin ^-1 \ \1/x = cosec^-1 x , x >= 1 ` or ` x <= -1`
• `cos^-1 \ \ 1/x = sec^-1 x, x ≥ 1` or `x ≤ – 1`
• `tan ^-1 \ \ 1/x = cot^-1 x , x > 0`
To prove the first result, we put `cosec^-1 x = y`, i.e., x = cosec y
Therefore `1/x = sin y`
Hence `sin^-1 \ \1/x = y`
or ` sin ^-1 \ \1/x = cosec^-1 x`
Similarly, we can prove the other parts.
`\color{red} \ox \color{red} (COMMON \ CONFUSION) `
`sin (sin^-1 x) = x, x ∈ [– 1, 1]` and `sin^-1 (sin x) = x, x ∈ [ - pi/2, pi/2]`
`cos (cos^-1 x) = x, x ∈ [– 1, 1]` and `cos ^-1 (cos x) = x, x ∈ [ 0,pi]`
`tan (tan^-1 x) = x, x ∈ R` and `tan^-1 (tan x) = x, x ∈ ( - pi/2, pi/2)`
`cosec (cosec^-1 x) = x, |x| ge 1` and `cosec^-1 (cosec x) = x, x ∈ [ - pi/2, pi/2] , xne0`
`sec (sec^-1 x) = x, |x| ge 1` and `sec^-1 (sec x) = x, x ∈ [ 0, pi], x ne pi/2`
`cot (cot^-1 x) = x, x ∈ R` and `cot^-1 (cot x) = x, x ∈ ( 0,pi)`
`\color{purple}ul(✓✓) \color{purple} "ALERT"`
`sin(sin^(-1)1) = 1`
`sin(sin^(-1)2) ne 2` (Firstly given question is wrong as `2` isn't in domain)
`\color { maroon} ® \color{maroon} ul (" REMEMBER") :` So if `sin(sin^(-1)x)` is given means `x` always will be in domain,So we can write directly equal to `x`
Property 1 :
• `sin ^-1 \ \1/x = cosec^-1 x , x >= 1 ` or ` x <= -1`
• `cos^-1 \ \ 1/x = sec^-1 x, x ≥ 1` or `x ≤ – 1`
• `tan ^-1 \ \ 1/x = cot^-1 x , x > 0`
To prove the first result, we put `cosec^-1 x = y`, i.e., x = cosec y
Therefore `1/x = sin y`
Hence `sin^-1 \ \1/x = y`
or ` sin ^-1 \ \1/x = cosec^-1 x`
Similarly, we can prove the other parts.