Find the principal value of: i) `sin^(-1)(-1/sqrt(2))`, ii) `cos^(-1)(-sqrt(3)/2)` iii) `tan^(-1)(-sqrt(3))` iv) `sec^(-1)(-2)` v) `"cosec"^(-1)(-sqrt(2))`, vii) `cot^(-1)(-1/sqrt(3))`.
Answered Feb 05, 2023
Correct Answer - i) `-pi/4`, ii) `(5pi)/6`, iii) `-pi/3`, iv) `(2pi)/3`, v) `-pi/4`, vi) `(2pi)/3`
Correct option is: A. cos θ sin θ
Correct Answer - C `AB=[(cos^(2) theta,),(cos theta sin theta,)][(cos^(2) phi,cos phi sin phi),(cos phi sin phi,sin^(2) phi)]` `=[(cos^(2) theta cos^(2) phi+cos theta cos phi sin theta sin phi ,cos^(2)theta cos phi...
(i) Since `1 in [-pi//2, pi//2], sin^(-1) (sin 1) = 1` (ii) Since `2 in [-pi//2, pi//2], sin^(-1) (sin 2) != 2` `:. Sin^(-1) (sin 2) = sin^(-1) (sin (pi)...
(i) Since `3 in [0, pi], cos^(-1) (cos 3) = 3` (ii) Since `4 !in [0, pi], cos^(-1) (cos 4) != 4` `:. Cos^(-1) (cos 4) = 2pi - 4`...
`E = sqrt(sin^(-1)x_(1)) sqrt(cos^(-1) x_(2)) + sqrt(sin^(-1) x_(2)) sqrt(cos^(-1) x_(3)) + sqrt(sin^(-1) x_(3)) sqrt(cos^(-1) x_(4)) +...+ sqrt(sin^(-1) x_(28)) sqrt(cos^(-1) x_(1))` `x_(i) in [0,1] AA i = 1, 2, 3, ..,...
Correct Answer - (a) `-(2pi)/(5)` (b) `(pi)/(3)` (c) `(2pi)/(5)` (d) `3 pi -8` (e) `10 - 3 pi` (f) `9 - 2pi` (g) `6 - pi` (h) `7 - 2pi` (a)...
Correct Answer - A Let `sqrt(tan alpha) = tan x`. Then `u = cot^(-1) (tan x) - tan^(-1) (tan x)` `= (pi)/(2) - x -x = (pi)/(2) - 2x` or `2x...
Correct Answer - B Let `x = sin theta and sqrtx = sin phi, " where " x in [0, 1]` `rArr theta, phi in [0, pi//2]` `rArr theta - phi...
Correct Answer - `[{:(,1,0),(,1,1):}]`
Correct Answer - i) `pi/3`, ii) `pi/6`, iii) `pi/3`, iv) `pi/4`, v) `pi/6`, vi) `pi/6`, vii) `pi/4`.
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