Question

sec 8 theta -1/sec 4 theta -1=tan 8 theta /tan 2theta​

Answer 1

Answer:

Trivially, sec8θ≥1sec8⁡θ≥1 and tan8θ≥0tan8⁡θ≥0 for all θ∈Rθ∈R for which secθsec⁡θ and tanθtan⁡θ are defined.

If a>0a>0 and b>0b>0. Then, sec8θa+tan8θb≥1a+0b=1a>1a+bsec8⁡θa+tan8⁡θb≥1a+0b=1a>1a+b, a contradiction.

If a<0a<0 and b<0b<0. Then, sec8θa+tan8θb