Equation `1+x^2+2x"sin"(cos^(-1)y)=0` is satisfied by exactly one value of `x` exactly two value of `x` exactly one value of `y` exactly two value of

By defing A & B are equal if they have the same order and all the corresponding elements are equal. Thus we have `sin theta=(1)/(sqrt2),c os theta=-(1)/(sqrt2)& tan theta=-1` `Rightarrow...

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...

Correct Answer - B We have. `[F(x) G(y)]^(-1)=[G(y)]^(-1) [F(x)]^(-1)` `=G(-y) F(-x)`

(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):}]`