Trig identity (1 Viewer)

[dasicmankev]

Hey, I just can't seem to prove this identity:

[1 + cosec^2 (A) tan^2 (C)] [1 + cot^2 (A) sin^2 (C)]
______________________ = ___________________

[1 + cosec^2 (B) tan^2 (C)] [1 + cot^2 (B) sin^2 (C)]

It'll be of great help if someone could solve this. Thanks.

 

[hscishard]

Have you tried dividing the numerator and denom by cosec^2 (B) tan^2 (C)?
If you haven't then idk.

 

[dasicmankev]

Yeah I tried but it wouldn't work unfortunately...

 

[AmieLea]

try the substitutions:

cosec^2(a)=1+cot^2(a)
and
tan^2(a)=sec^2(a)-1

i've tried and it looks like it's going to work, but i just haven't quite got there yet

tell us if it works, or if you figured it out. i love this kind of stuff, but this one has me stumped lol. i'l keep trying

 

[dasicmankev]

Thanks guys I got it.

LHS = 1 + cosec^2 (A) tan^2 (C)
______________________

1 + cosec^2 (B) tan^2(C)


= 1 + [1 + cot^2 (A)] tan^2 (C)
_________________________

1 + [1 + cot^2 (B)] tan^2 (C)


= 1 + tan^2 (C) + cot^2 (A) tan^2 (C)
_______________________________

1 + tan^2 (C) + cot^2 (B) tan^2 (C)


= sec^2 (C) + cot^2 (A) tan^2 (C)
_____________________________

sec^2 (C) + cot^2 (B) tan^2 (C)


= cos^2 (C) sec^2 (C) + cot^2 (A) sin^2 (C)
____________________________________

cos^2 (C) sec^2 (C) + cot^2 (B) sin^2 (C)


= 1 + cot^2 (A) sin^2 (C)
____________________

1 + cot^2 (B) sin^2 (C)

= RHS

 

[hscishard]

Nice work.