Tan^2/sec^2 + cot^2/csc^2 = 1...please prove?

Question

Tan^2/sec^2 + cot^2/csc^2 = 1...please prove?

Answer 1

sec^2=1/cos^2 csc^2=1/sin^2

replace all this

tan^2 * cos^2 + cot^2 *sin^2 = sin^2 + cos^2 =1

Answer 2

O is the opposite side to angle x in a right angled triangle, A is adjacent and H is hypotenuse.  Then, from the definitions of the trig functions, your expression is (O^2/A^2 x A^2/H^2 )+ (A^2/O^2 x O^2/H^2) which reduces to(O^2/H^2 + (A^2/H^2) which is sin^2 + cos^2 which according to Pythagoras'Theorem=1.

Answer 3

sin^2/cos^2 / 1/cos^2 + cos^2/sin^2/1/sin^2

sin^2 + cos^2 =1

Answer 4

tan^2= sin^2/cos^2           sec^2= 1/cos^2

cot^2 = cos^2/sin^2          csc^2= 1/sin^2

so tan^2/sec^2 = (sin^2/cos^2)/ (1/cos^2) = sin^2

    cot^2/csc^2  = (cos^2/sin^2) / (1/sin^2) = cos^2

and sin^2+cos^2= 1 so the problem is proved