If `theta-phi=pi/2,` prove that, `[(cos^2 theta,cos theta sin theta),(cos theta sin theta,sin^2 theta)] [(cos^2 phi,cos phi sin phi),(cos phi sin phi,

`cos^(-1).(x)/(a) + cos theta +(y^(2))/(b^(2)) = sin^(2) theta` `rArr cos^(-1)[(x)/(a).(y)/(b)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))] = theta` `rArr (xy)/(ab) -sqrt(1-(x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(2)y^(2))/(a^(2)b^(2)))= cos theta` `rArr " " (xy)/(ab)-costheta =sqrt(1-(x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(3)y^(2))/(a^(2)b^(2)))` squaring both sideswe get `(x^(2)y^(2))/(a^(2)b^(2))+ cos^(2) theta -(2xy)/(ab)...

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`

Correct Answer - A More are the number of `alpha-`hydrogen present in the alkyl group attached to the benzene ring more pronounced will be the hyperconjugation and the benzene ring will...

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`...

We have, `|A-lambdaI|=|(cos theta - lambda,-sin theta),(sin theta,cos theta - lambda)|` `=(cos theta-lambda)^(2)+sin^(2) theta` Therefore, characteristic equation of A is `(cos theta-lambda)^(2)+sin^(2) theta=0` or `cos theta-lambda= pm i sin theta`...

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...

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):}]`

Correct Answer - B::C All are infinte geometric progression with common ratio lt 1 `x=1/(1-cos^2phi)=1/sin^2phi,y=1/(1-sin^2phi)=1/cos^2phi`, `z=1/(1-cos^2phisin^2phi)` Now, `xy+z=1/(sin^2phicos^2phi)+1/(1-sin^2phicos^2phi)` `=1/(sin^2phicos^2phi(1-sin^2phicos^2phi))` `or xy+z=xyz ...(i)` Clearly, `x+y=(sin^2phi+cos^2phi)/(sin^2phicos^2phi)=xy` `:. x+y+z=xyz` [using Eq. (i)]