(i) `sec(cos^(-1)""(1)/(2))` का मान ज्ञात कीजिए।
(ii) समीकरणों `sin^(-1)x+sin^(-1)y=(2pi)/(3)` और `cos^(-1)x-cos^(-1)y=(pi)/(3)` को हल कीजिए।
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...
2 Answers 1 views(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)...
2 Answers 1 views(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`...
2 Answers 1 viewsCorrect 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)...
2 Answers 1 viewsCorrect Answer - C `sin^(-1) (sin theta) gt (pi)/(2) - sin^(-1) (sin theta)` `rArr sin^(-1) (sin theta) gt (pi)/(4)` `rArr sin theta gt (1)/(sqrt2)` `rArr (pi)/(4) lt theta lt (3pi)/(4)`
2 Answers 1 viewsCorrect Answer - B `sin^(-1) x in [-(pi)/(2), (pi)/(2)]` `cos^(-1) y in [0, pi]` `sec^(-1) z in [0, (pi)/(2)) uu ((pi)/(2), pi]` `rArr sin^(-1) x + cos^(-1) y + sec^(-1) z...
2 Answers 1 viewsCorrect Answer - C `sin^(-1) x in [-(pi)/(2), (pi)/(2)]` `cos^(-1) y in [0, pi]` `sec^(-1) z in [0, (pi)/(2)) uu ((pi)/(2), pi]` `rArr sin^(-1) x + cos^(-1) y + sec^(-1) z...
2 Answers 1 viewsCorrect Answer - `[{:(,1,0),(,1,1):}]`
2 Answers 1 views`y=sqrt((1-sin 2x)/(1+ sin 2x))=(cos x - sin x)/(cos x + sin x)=(1-tan x)/(1+ tan x)=tan ((pi)/(4)-x)` `therefore (dy)/(dx)=-sec^(2)((pi)/(4)-x)`
2 Answers 1 viewsCorrect Answer - `{:((a)(1)/(2),(b)(1)/(sqrt(2)),(c)sqrt(3)),((d)0,(e)(1)/(2),(f)(1)/(2)):}`
2 Answers 1 views