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

Prove that (sin^(2)theta+cos^(2)theta)/(sec^(2)theta-tan^(2)theta)=1

Prove that (sin^(2)30^(@)+cos^(2)30^(@))/(sec^(2)57^(@)-tan^(2)57^(@))=1

The expression cosec^(2)A cot^(2)A-sec^(2)A tan^(2)A-(cot^(2)A-tan^(2)A)(sec^(2)A cosec^(2)A-1) is equal to

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

Without using trigonometric tables , prove that : (i) sin50^(@)-cos40^(@)=0 (ii) tan36^(@)-cot54^(@)=0 (iii) sec25^(@)-"cosec "65^(@)=0 (iv) sin^(2)32^(@)+sin^(2)58^(@)=1 (v) "cosec"^(2)39^(@)-tan^(2)51^(@)=1 (vi) sec^(2)10^(@)-cot^(2)80^(@)=1

If A=60^(@) , verify that : (i) sin^(2)A+cos^(2)A=1" (ii) "sec^(2)A-tan^(2)A=1

If A=60^(@) , verify that : (i) sin^(2)A+cos^(2)A=1" (ii) "sec^(2)A-tan^(2)A=1

Show that (i) sin^(8)A-cos^(8)A=(sin^(2)A-cos^(2)A)(1-2sin^(2)A.cos^(2)A) (ii) (1)/(sec A-tan A)-(1)/(cos A)=(1)/(cos A)-(1)/(sec A + tan A)

Without using trigonometric tables, evaluate each of the following: (cos^2 20^0+cos^2 70^0)/(sec^2 50^0-cot^2 40^0)+2cosec^2 58^0-2cot58^0tan32^0- 4tan13^0tan37^0tan45^0tan53^0tan77^0

Prove that : (i) tan^(2)A-sin^(2)A=tan^(2)A*sin^(2)A (ii) cot^(2)theta-cos^(2)theta=cot^(2)theta*cos^(2)theta .