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Consider the following statements.
I. $\sin ^{-1}\left(y^2-4 y+6\right)+\cos ^{-1}\left(y^2-4 y+6\right) =\frac{\pi}{2}, \forall y \in R$
II. $\sec ^{-1}\left(y^2-4 y+6\right)+\operatorname{cosec}^{-1}\left(y^2-4 y+6\right) =\frac{\pi}{2}, \forall y \in R$
Which of the above statement(s) is/are true?
Options:
I. $\sin ^{-1}\left(y^2-4 y+6\right)+\cos ^{-1}\left(y^2-4 y+6\right) =\frac{\pi}{2}, \forall y \in R$
II. $\sec ^{-1}\left(y^2-4 y+6\right)+\operatorname{cosec}^{-1}\left(y^2-4 y+6\right) =\frac{\pi}{2}, \forall y \in R$
Which of the above statement(s) is/are true?
Solution:
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Verified Answer
The correct answer is:
Only II
Given statements,
$\sin ^{-1}\left(y^2-4 y+6\right)+\cos ^{-1}\left(y^2-4 y+6\right)=\frac{\pi}{2}$
$\because y^2-4 y+6=(y-2)^2+2 \geq 2, \forall y \in R \text {. }$
So, statement (i) is false and statement (ii) is true.
$\sin ^{-1}\left(y^2-4 y+6\right)+\cos ^{-1}\left(y^2-4 y+6\right)=\frac{\pi}{2}$
$\because y^2-4 y+6=(y-2)^2+2 \geq 2, \forall y \in R \text {. }$
So, statement (i) is false and statement (ii) is true.
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