If sin\( \theta \) = K find tan\(\theta\), 0° \(\leq\) \(\theta\) \(\leq\) 90°.
1-K
\( \frac{k}{k - 1} \)
\( \frac{k}{\sqrt{1 - k^2}} \)
\( \frac{k}{1 - k} \)
\( \frac{k}{\sqrt{ k^2 - 1}} \)
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Okay guys am back so it seems like the question isn't clear for a lot of people. So, lets go straight to it
So Sin θ = K
we all know that
Sin θ = Opposite/ Hypothenus
Sin θ = K/1
i.e hypothenus is 1
using Pythagorus theorem: Hypothenusθ² = opposite = adjacent
Lets call
Hypothenus - Hyp
Opposite - Opp
Adjacent - Adj
Hyp = 1
Opposite = K
Adjacent = ? Lets call it X
*(Using the pythagorus theorem)*
1² = K² + X²
1 = K² + X²
*(making X the subject of the formula)*
X² = 1 - K²
X = √(1 - K²)
Since, X = Cosθ
Cosθ = √(1 - K²)
Remember Tanθ = Sinθ/Cosθ
Tanθ = K / √(1 - K²)
Hop eyou all understand. Thank you 
Wow surprising 😮😹, sin(∅⁰)=opp/hyp, but they gave the value of opp and the hypotenus then reference to Pythagoras theorem which is hpy²=opp²+adj² (input this values opp=k, hyp=1, which gives adj²=1²-k² i.e adj=√(1-k²) then put it into tan(∅⁰)=opp/adj==>tán(∅⁰)=1/(√1-k²), hope this is clear enough?
