Question

tan^4 h -1 I need to express this in terms of sin h and cos h and simplify. I typed it wrong earlier. Sorry.

Answer 1

is it really\[tan^4(h-1)\] because if so this is a silly problem as tan is just sine over cosine. perhaps
\[tan^4(h)-1\]?

Answer 2

I don't know. This is how the teacher typed it on her worksheet. Could you show me how to simplify tan^4(h)-1. Thanks

Answer 3

\[tan^4(h)-1=(tan^2-1)(tan^2+1)\]
and
\[tan^2(h)+1=sec^2(h)\]
\[(tan^2(h)-1)sec^2(h)=\frac{tan^2(h)-1}{cos^2(h)}\]
\[=\frac{\frac{sin^2(h)}{cos^2(h)}-1}{cos^2(h)}\]
\[=\frac{sin^2(h)}{cos^4(h)}-\frac{1}{cos^2(h)}\]
or if you prefer
\[\frac{sin^2(h)-cos^2(h)}{cos^4(h)}\]