Question

Prove that sec²(90-theta) - 1/( cot²(90- theta) =1

Answer 1

Answer:

LHS = sec^2 ( 90 - theta) - 1/(cot^2(90- theta)

we know that

sec^2 (90- theta) = cosec^2

cot^2(90- theta) = tan^2

cosec^2 - 1/ tan^2

1/ sin^2 - 1/ (sin/cos)^2

1/sin^2 - cos^2/sin^2

find LCM between sin^2 and sin^2

LCM between sin^2 and sin^2 is sin^2

1 - cos^2 /sin^2

sin^2/sin^2 = 1

LHS = RHS

Answer 2

Answer:

Answer:

LHS = sec^2 ( 90 - theta) - 1/(cot^2(90- theta)

we know that

sec^2 (90- theta) = cosec^2

cot^2(90- theta) = tan^2

cosec^2 - 1/ tan^2

1/ sin^2 - 1/ (sin/cos)^2

1/sin^2 - cos^2/sin^2

find LCM between sin^2 and sin^2

LCM between sin^2 and sin^2 is sin^2

1 - cos^2 /sin^2

sin^2/sin^2 = 1

LHS = RHS