if `theta=(11pi)/6` then `costheta+sintheta=`

If : cos theta-sintheta=sqrt2.sin theta, "then": costheta+sintheta=

costheta[[costheta, sintheta], [-sintheta, costheta]]+sintheta[[sintheta, -costheta], [costheta, sintheta]]=

Find (d^(2)y)/(dx^(2))" at "theta=pi/2 when : x=a(theta+sintheta),y=a(1-costheta)

Find (d^(2)y)/(dx^(2))" at "theta=pi/2 when : x=a(theta-sintheta),y=a(1-costheta)

Find (d^(2)y)/(dx^(2))" at "theta=pi/2 when : x=a(theta-sintheta),y=a(1+costheta)

Eliminate theta in the following (i) 2sintheta=x, 3costheta=y (ii) costheta + sintheta=x, costheta-sintheta=y

For any acute angle theta; Prove (i) 'tan theta = sintheta/costheta (ii) cot theta =cos theta/sintheta

If costheta +sin theta= sqrt(2)cos theta , then prove that costheta-sintheta =sqrt(2)sin theta

Simplify: cos theta[{:(costheta,sintheta),(-sintheta,costheta):}]+sintheta[{:(sin theta ,-costheta),(costheta, sintheta):}]

costheta[{:(costheta,-sin theta),(sintheta,costheta):}]+sintheta[{:(sintheta,costheta),(-costheta,sintheta):}]=?