If asin²A+bcos²A=c. Prove that cot²A+c-a/b-c​

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

the correct answer is a) "catholic church".

the proposals that liberals in mexico made during 1855 were particularly focused on a person's individual rights, such as the freedom of speech, the freedom of conscience, freedom of assembly, and the right to possess weapons.

when the constitution was formally published 2 years later (1857). it also contained an article which permitted education free from any religious influence, the removal of the catholic church institutional privileges, and the sale of property belonging to the church. all of these provisions were clearly against the catholic church interests.