If `cos ^(-1) .(x)/(a)+ cos ^(-1). (y)/(b) = theta` then prove that `(x^(2))/(a^(2)) -(2xy)/(ab) . cos theta+ (y^(2))/(b^(2)) = sin ^(2) theta`.

Correct option is: B) Sin θ Cos θ \(Sin^3 \theta \, Cos \theta + Cos^3 \theta \, Sin \theta =\) sin \(\theta\) cos \(\theta\) (\(sin^2\theta + cos^2\theta\)) =  sin \(\theta\) .cos \(\theta\)  (\(sin^2\theta + cos^2\theta\) = 1)

Given that `sin theta=1-sin^(2)theta=cos^(2)theta` LHS= `cos^(6)theta(ccos^(2)theta+1)^(3)-1=sin^(3)theta(1 +sintheta)^(3)-1=(sin theta+sin^(2)theta)^(3)-1=1-1=0`

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 - A::B::C We have, `|A(theta)|=1` Hence, A is invertiable. `A(pi+theta)=A(pi + theta)=[(sin(pi+theta),i cos (pi+theta)),(i cos (pi+theta),sin (pi+theta))]` `=[(-sin theta,-i co theta),(-i cos theta,-sin theta)]=-A(theta)` adj `(A(theta))=[(sin theta,-i cos theta),(-i...

(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)...

Correct Answer - 9 We must have `4 sin^(2) theta + sin theta = 6 sin theta -1` `rArr 4 sin^(2) theta - 5 sin theta + 1 = 0` `rArr...

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