If `F(x)=[("cos"x,-sin x,0),(sin x,cos x,0),(0,0,1)]` and `G(y)=[(cos y,0,sin y),(0,1,0),(-sin y,0,cos y)]`, then `[F(x) G(y)]^(-1)` is equal to

Let matrix `A=[(a,b),(c,d)]` is orthogonal matrix. `:. [(a,b),(c,d)][(a,c),(b,d)]=[(1,0),(0,1)]` `implies [(a^(2)+b^(2),ac+bd),(ac+bd,c^(2)+d^(2))]=[(1,0),(0,1)]` `implies a^(2)+b^(2)=1` (1) `c^(2)+d^(2)=1` (2) `ac+bd=0` (3) `implies a/d= (-b)/c= k` (let) `implies c^(2)+d^(2)=(a^(2)+b^(2))/k=1//k^(2)` or `k^(2)=1` or `k= pm 1`...

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`...

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 - C `a = 20 - 6pi, b = 10 pi - 30, c = sin^(-1) sin(4pi - 10) = 10 - 3pi` So, `a + b + c...

Correct Answer - 9 `1 + sin (cos^(-1) x) + sin^(2) (cos^(-1) x) + ...oo = 2` `rArr (1)/(1-sin (cos^(-1)x)) = 2` or `(1)/(2) = 1 - sin (cos^(-1) x)` or...

Correct Answer - `[{:(,1,0),(,1,1):}]`

Correct Answer - B `L.H.S. = (1 + cos A)/(a) + (1 + cos B)/(b) + (1 + cos C)/(c)` `= (2 cos^(2).(A)/(2))/(2R sin A) + (2 cos^(2).(B)/(2))/(2R sin B) +...

Correct Answer - A `(a)` `a=e^(ialpha)`, `b=e^(ibeta)`, `c=e^(igamma)` Clearly `a+b+c=0` `impliesa+b+c=0=(1)/(a)+(1)/(b)+(1)/(c )` `impliesa^(3)+b^(3)+x^(3)=3abc` `impliessin3alpha+sin3beta+sin3gamma=sin(alpha+beta+gamma)`