3.
Prove that tan (cot-' x) = cot (tan-'x).
Answers
Step-by-step explanation:
sin (sin-1 x) = x, x ∈ [–1, 1]
Let sin-1 x = θ, θ &isin (-π/2,π/2) ……… (1)
Then x = sin θ
From (1) putting the value of θ in (2), we get, ……… (2)
sin (sin-1 x) = sin (θ) = x
(b) sin-1 sin x = x, x ∈ [-π/2,π/2].
Let sin x = y, y ∈ [–1, 1] ……… (1)
Then x = sin-1 y ……… (2)
Putting the value of y in (2) from (1), we get,
sin-1 sin x = x
Illustration:
Evaluate sin-1 sin 5, 5 is radian
Solution:
We know sin-1 sin x = x, x ∈ [-π/2,π/2] or x ∈ [–1.57, 1.57]
We can write, 5 = 2 π + 5 – 2 π
sin 5 = sin (2 π + (5 – 2π))
= sin (5 – 2 π), 5 – 2 π ∈ [–1.57, 1.57]
∴ sin-1 sin 5 = sin-1 sin (5 – 2π)
= 5 – 2π
2. Reciprocal Property
sin-1 (1/x) = cosec-1 x, x ≤ –1, or x ≥ 1
let cosec-1 x = θ ⇒ cosec θ, θ ∈ [-π/2,π/2] – {0}
⇒ LHS = sin-1 (1/(cosec θ )) = θ
⇒ sin-1 (sin θ) = θ which is true for above domain of θ
= cosec-1 x = RHS.
Note: cot-1 (1/x) = tan-1 x holds only for x > 0. This is because of principal value range of Inverse function cot and tan are different.
When x < 0, the principal value range is π/2 < cot-1 (1/x) < π, whereas -π/2 < tan-1x< 0. Hence equality will never hold in this case.
cot-1 (-1/2) = tan-1 (–x)
π – cot-1 (1/2) = –tan-1 x
cot-1 (1/2) = π + tan-1 x
Hence cot-1 (1/2) = { tan-1 x, x>0
π+tan-1 x, x< 0
Relation between Inverse Trigonometric Functions
1. sin-1 x + cos-1 x = π/2 x ∈ [–1, 1]
Let sin-1 x = θ, θ ∈ [-π/2,π/2]
⇒ x = sin θ,
⇒ x = cos (π/2-θ ),(π/2-θ ) ∈ [0, π]
⇒ cos-1 x = π/2 – θ
⇒ cos-1 x + sin-1 x = π/2 x ∈ [–1, 1].
Answer:
hi friends
Step-by-step explanation:
sin- 1×+ cos-1×=π/2×€[-1,1]
let sin-1×=8,8€[-π/2,π/2]
=X=sin 8
=X=cos(π/2-8),(π/2-8)€[0,π]
=cos-1×=π/2-8
=cos-1×+sin-1×=π/2×€[-1,1]
I hope this will help you