Prove that ` |(1,x,1),(0,cos x, sin y),(0 , sin x, cos y )|=cos (x+y)`

Explanation

(a) The equation of tangent at `((cos theta)//2,(sin theta)//3) "is" 2x cos theta+3ysin theta=1`
Which is parallel to the given line `8x=9y`. Therefore, `cos theta=+-(4)/(5), sin =(+-3)/(5)`
Hence, the points are (2/5,-1/5) and (-2/51/5).
The distance between the points is `sqrt((16)/(25)+(4)/(25))=(2)/(sqrt(5))`
Which is the given
The given equation is `((x+1)^(2))/(9)+((y+2)^(2))/(25)=1`
or `e^(2)=1-(9)/(25)=(16)/(25)or e=(4)/(5)`
Hence, the foci, `S,S-=(1,-2+-4)-=S(-1,2) and S(-1.-6)`
Required sum of distances `=2+6=8`
(c) The equation of normal at `(3sos theta, 2 sin theta)` s `3xsec theta-2y "cosec" theta=5`
whis is parallel to the given line `2x+y=1`. Therefore,
`cos theta=+-(3)(5), sin theta=+-(4)/(5)`
Hence, the points are `(-9/5,9/5) and (9/5,-8/5)`
Required sum of the distances`=(16)/(5)`
(d) Conider any point `(t,t+2), t in R` on the line `x-y+2=0`
The chord of contact of the ellipse w.r.t. this point is
`xt+2y(t+2)=2`
or `(4y-2)+(x+2y)=0`
This line passes trough the points of the intersection of lines
`4y-2=0 and x+2y=0`
`:. x=-1,y=1//2`
Hence, the point is `(-1,1//2)` os 3.