Solve: Write the product of two trig equations that is equal to one.
This one confuses me because I can think of trig equations that equal one, but I can't think of trig equations that I could multiply times eachother in order to equal one.
Or maybe I'm missing something?
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$\begingroup$how about $$(\sin^2x+\cos^2x)(\sec^2x-\tan^2x)=1\times1=1$$ or, using the second bracket as the difference of two squares, $$\sec^2x-\tan^2x=(\sec x-\tan x)(\sec x+\tan x)=1$$
$\endgroup$ $\begingroup$$$(\sin^2x+\cos^2x)(\csc^2x-\cot^2x)=1\cdot1=1$$
$\endgroup$ $\begingroup$It need not be only algebraic, many reciprocal Trig relations exist, among them:
Trig
$$(\sec x-\tan x)(\sec x+\tan x)=1$$
$$ \tan ( 45^\circ -A ) \tan (45^\circ + A) = 1 $$
Hyperbolic
$$(\cosh x+\sinh x) (\cosh x-\sinh x) =1 $$
Complex Trig
$$(\cos x+ i \sin x) (\cos x- i \sin x) =1 $$
Geometrical
If a cone of semi vertical angle $\alpha$ is cut by a plane inclined at $\beta$ to axis then
$$ cos\alpha \cdot \sec \beta = 1 $$ and so on.
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