c o s − 1 x = t a n − 1 √ 1 − x 2 x , then: Options: A . &nbsp X ∈ R B . &nbsp − ≤ x ≤ 1 , x ≠ 0 C . &nbsp 0 < x ≤ 1 D . &nbsp N o n e o f t h e s e

Answer: Option C : C Putting θ = c o s − 1 x in R.H.S., we have R.H.S = t a n − 1 √ 1 − x 2 x = t a n − 1 √ 1 − c o s 2 θ c o s θ ( 0 ≤ θ ≤ π ) = t a n − 1 s i n θ c o s θ = t a n − 1 t a n θ = θ = c o s − 1 x when − π 2 < θ < π 2 i.e., when 0 ≤ θ < π 2 i.e., when 0 < c o s θ ≤ 1 i.e., 0 < x ≤ 1 .

cos−1x=tan−1√1−x2x, then:

Discussion Forum : Inverse Trigonometric Functions

Question -

$c o s^{- 1} x = t a n^{- 1} \frac{\sqrt{1 - x^{2}}}{x}$ , then:

Options:

A .

X \in R

B .

- \leq x \leq 1, x \neq 0

C .

0 < x \leq 1

D .

N o n e o f t h e s e

Answer: Option C
:
C
Putting

θ = c o s^{- 1} x

in R.H.S., we have
R.H.S

= t a n^{- 1} \frac{\sqrt{1 - x^{2}}}{x} = t a n^{- 1} \frac{\sqrt{1 - c o s^{2} θ}}{c o s θ} (0 \leq θ \leq π) = t a n^{- 1} \frac{s i n θ}{c o s θ} = t a n^{- 1} t a n θ = θ = c o s^{- 1} x

when

- \frac{π}{2} < θ < \frac{π}{2}

i.e., when

0 \leq θ < \frac{π}{2}

i.e., when

0 < c o s θ \leq 1

i.e.,

0 < x \leq 1

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Discussion Forum : Inverse Trigonometric Functions

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