InterviewSolution
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InterviewSolution
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If `sin^(-1) : [-1, 1] rarr [(pi)/(2), (3pi)/(2)] and cos^(-1) : [-1, 1] rarr [0, pi]` be two bijective functions, respectively inverse of bijective functions `sin : [(pi)/(2), (3pi)/(2)] rarr [-1, 10 and cos : [0, pi] rarr [-1, 1] " then " sin^(-1) x + cos^(-1) x` isA. `(pi)/(2)`B. `pi`C. `(3pi)/(2)`D. not a constant |
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Answer» Correct Answer - D Let `sin^(-1) x = theta` `rArr x = sin theta, (pi)/(2) le theta le (3pi)/(2)` Now, `cos^(-1) x = cos^(-1) (sin theta)` `= cos^(-1) (-cos ((3pi)/(2) - theta))` `= pi - cos^(-1) (cos ((3pi)/(2) - theta))` `= pi -((3pi)/(2) - theta), " as " 0 le (3pi)/(2) - theta le pi` `= theta -(pi)/(2) = sin^(-1) x - (pi)/(2)` Hence, `sin^(-1) x + cos^(-1) x = 2 sin^(-1) x - (pi)/(2)` |
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