(1+tan^2A)/(1+cot^2A)=((1-tanA)^2)/(1-cotA)^2
3 answers
Yes, this equation is true.
ok - I think the point of the exercise, though, is to prove it.
Stupid bot
Left Side:
(1+tan^2A)/(1+cot^2A)
= sec^2A/csc^2A
= tan^2A
Right Side:
(1-tanA)^2/(1-cotA)^2
= (1 - sinA/cosA)^2/(1 - cosA/sinA)^2
= ((cosA-sinA)/cosA)^2 / ((sinA-cosA)/sinA)^2
= sin^2A/cos^2A
= tan^2A
QED
Stupid bot
Left Side:
(1+tan^2A)/(1+cot^2A)
= sec^2A/csc^2A
= tan^2A
Right Side:
(1-tanA)^2/(1-cotA)^2
= (1 - sinA/cosA)^2/(1 - cosA/sinA)^2
= ((cosA-sinA)/cosA)^2 / ((sinA-cosA)/sinA)^2
= sin^2A/cos^2A
= tan^2A
QED
The question probably was to "prove it", not to agree or disagree with the
statement.
LS = (1+tan^2A)/(1+cot^2A)
= sec^2 A/csc^2 A)
= sin^2 A/cos^2 A = tan^2 A
RS = ((1-tanA)^2)/(1-cotA)^2
= ( (1 - tanA)/(1 - cotA) )^2
= [ ( 1 - sinA/cosA)/(1 - cosA/sinA) ]^2
= [ ((cosA - sinA)/cosA)/((sinA - cosA)/sinA) ]^2
= [ - sinA/cosA]^2
= [-tanA]^2
= tan^2 A
= LS
statement.
LS = (1+tan^2A)/(1+cot^2A)
= sec^2 A/csc^2 A)
= sin^2 A/cos^2 A = tan^2 A
RS = ((1-tanA)^2)/(1-cotA)^2
= ( (1 - tanA)/(1 - cotA) )^2
= [ ( 1 - sinA/cosA)/(1 - cosA/sinA) ]^2
= [ ((cosA - sinA)/cosA)/((sinA - cosA)/sinA) ]^2
= [ - sinA/cosA]^2
= [-tanA]^2
= tan^2 A
= LS