Evaluate `(sin^(2)2020+cos^(2)2020+tan50)/(sec^(2)2019-tan^(2)2019+cot40)=`

Evaluate : (sec^(2)theta-cot^(2)(90^(@)-theta))/("cosec"^(2)67^(@)-tan^(2)23^(2))+(sin^(2)40^(@)+sin^(2)50^(@))

Prove :sin^(2)A cot^(2)A+cos^(2)A tan^(2)A=1

Prove that (sec^(2)A)/(cos^(2)A)-(tan^(2)A)/(cot^(2)A)=1+2tan^(2)A

(sec^2A)/(cos^2A)-(tan^2A)/(cot^2A)=1+2tan^2A

Evaluate : (2sin68^(0))/(cos22^(0))-(2cot15^(0))/(5tan75^(0))-(3tan45^(0)tan20^(@)tan40^(@)tan50^(@)tan70^(@))/(5)

tan^(2)A+cot^(2)A=sec^(2)A cos ec^(2)A-2

sin ^ (2) A.cot ^ (2) A-cos ^ (2) A * tan ^ (2) A = 1

Evaluate : (3tan25^(@)tan40^(@)tan50^(@)tan65^(@)-(1)/(2)tan^(2)60^(@))/(4(cos^(2)29^(@)+cos^(2)61^(2)))

Without using trigonometric tables , evaluate : (i) (sin40^(@))/(cos50^(@))+(3tan50^(@))/(cot40^(@)) (ii) (sec37^(@))/("cosec "53^(@))-(tan20^(@))/(cot70^(@)) (iii) (cos74^(@))/(sin16^(@))+(sin12^(@))/(cos78^(@))-sin18^(@)sec72^(@) (iv) sin35^(@)sec55^(@)+(4sec32^(@))/("cosec "58^(@))

(sin A+cosec A)^(2)+(cos A+sec A)^(2)=7+tan^(2)A+cot^(2)A