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There no rule of trig that says it’s true.

Perhaps for *this particular* right triangle, you are being given extra information beyond standard rules?
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It isn't.
by
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Are you winding us up, OP? :)
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Assume this is a right triangle, and that the hypotenuse has length *f*. Let *a* be the opposite sidelength, and *b* the adjacent. We have that sinα = *a*/*f*, and cosα = *b*/*f*. We further have that tanα = *a*/*b* = (*f* sinα)/(*f* cosα). That is, the only way tanα = *f*, is when sinα/cosα = *f*, or equivalently sinα = *f* cosα.     
-1 <= sinα <= 1, and same holds for cosine. This forces your *f* = 1, and your angle to α = 45°.
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I don't see why that should be true

Use SOHCAHTOA
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Is this a right triangle even?
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If tan(alpha)=f, and f²=y²+x², and tan(alpha)=y/x then y²+x²=tan²(alpha)=y²/x².  
y²x²+x⁴=y²  
y²(x²-1)=-x⁴  
y²=x⁴/(-x²+1)  
Assuming y and x are real-valued and positive, this means x² must be between 0 and 1 (and therefore x is,  as well), but there are no particular constraints, beyond that. y is a function of x, and it's one-to-one for 0<x<1, so as your triangle gets taller, the x side must get closer and closer to 1, asymptotically.
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Tan(a) = opposite/adjacent
ago
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I’ll just add that tan alpha = f doesn’t make sense physically since f has units of length. But sure, if length is just a number you could rescale any right angles triangle, keeping alpha the same, until the hypotenuse is tan alpha or cos alpha or any positive number you want.
ago
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Suppose the adyacent catetus is a and opposite catetus is b. Then, by definition of tan(alpha), we have f=a/b (*) and, by Pythagoras, we have also a² + b² = f² (**). Then wey have have a equation's system which incognites ate a and b.

f=a/b ( *)

a²+b²=f² (**)


By ( *) wey have a=f•b (***)


Then, reemplacing ( ***) in (**) and solving for b we have


(f•b)² + b² = f²

f² • b²  + b² = f²

(f² + 1) • b² = f²

b² = f² / (f² + 1)

b = (f² / (f² + 1))^(1/2)

b = f /((f² + 1))^(1/2)

b = f • (f² + 1)^(1/2) / (f² + 1). (****)

Then, reemplacing (****) in (***) a solving for a we have

a=b•f

a = (f • (f² + 1)^(1/2) / (f² + 1)) • f

a = f² • (f² + 1)^(1/2) / (f² + 1)
ago

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