Question

sec(tan^-1 u + cos^-1 v) sec(tan^-1 u + cos^-1 v) @Mathematics

Answer 1

ick.
use addition angle formula for cosine and then take the reciprocal

Answer 2

oh, I just posted on this the second time you asked (@physics)

Answer 3

Here's a copy
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well, cos(x + y) = cos x . cos y - sin x . sin y

Hence
cos(tan^-1 u + cos^-1 v) = cos(arctan u).v - sin(arctan u).sin(arccos v)

You should be able to reduce that expression further now (if you're stuck, draw a triangle). Then take the reciprocal to get back to sec.

Answer 4

\[\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\] with
\[a=\tan^{-1}(u)\]

\[b=\cos^{-1}(v)\]
so you need
\[\cos(\tan^{-1}(u))=\frac{1}{\sqrt{u^2+1}}\]

\[\cos(\cos^{-1}(v))=v\]
\[
sin(\tan^{-1}(u))=\frac{u}{\sqrt{u^2+1}}\]
\[\sin(\cos^{-1}(v))=\sqrt{1-v^2}\] and plug all that mess in

Answer 5

what jamesj said. took me like ten minutes to type this!