If sinθθ = K find tanθθ, 0° ≤≤ θθ ≤≤ 90°.

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1. 

   Dr Ben June 30, 2025

   You’re given:
   $\sin \left(\theta \theta \right)=K\backslash sin(\theta \theta )\; =\; K$sin(θθ)=K
   and asked to find:
   $\tan \left(\theta \theta \right)\backslash tan(\theta \theta )$tan(θθ)
   for $0^{\circ }\le \theta \theta \le {90}^{\circ }0^\backslash circ\; \backslash leq\; \theta \theta \; \backslash leq\; 90^\backslash circ$0∘≤θθ≤90∘.
   Assuming the equation means:

   • $\sin \left(\theta \theta \right)=K\backslash sin(\theta \theta )\; =\; K$sin(θθ)=K (i.e., “θθ” is treated as a single angle, so it’s not a typo of $\sin \theta \cdot \theta \backslash sin\theta \; \backslash cdot\; \theta$sinθ⋅θ)
   • Then we want $\tan \left(\theta \theta \right)\backslash tan(\theta \theta )$tan(θθ)

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   Step-by-step:

   From:
   $\sin \left(\theta \theta \right)=K\backslash sin(\theta \theta )\; =\; K$sin(θθ)=K
   Then:
   $\theta \theta ={\sin }^{-1}\left(K\right)\theta \theta \; =\; \backslash sin^\{-1\}(K)$θθ=sin−1(K)
   Then:
   $\tan \left(\theta \theta \right)=\tan \left({\sin }^{-1}\right(K\left)\right)\backslash tan(\theta \theta )\; =\; \backslash tan(\backslash sin^\{-1\}(K))$tan(θθ)=tan(sin−1(K))

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   Identity:

   Let $x={\sin }^{-1}\left(K\right)x\; =\; \backslash sin^\{-1\}(K)$x=sin−1(K). Then:
   $\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}=\frac{K}{\sqrt{1-K^2}}\backslash tan(x)\; =\; \backslash frac\{\backslash sin(x)\}\{\backslash cos(x)\}\; =\; \backslash frac\{K\}\{\backslash sqrt\{1\; –\; K^2\}\}$tan(x)=cos(x)sin(x)​=1−K2​K​

   (From the Pythagorean identity $\cos \left(x\right)=\sqrt{1-{\sin }^2\left(x\right)}\backslash cos(x)\; =\; \backslash sqrt\{1\; –\; \backslash sin^2(x)\}$cos(x)=1−sin2(x)​, valid in the range $0^{\circ }\le x\le {90}^{\circ }0^\backslash circ\; \backslash le\; x\; \backslash le\; 90^\backslash circ$0∘≤x≤90∘).

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   Final Answer:

   $\tan \left(\theta \theta \right)=\frac{K}{\sqrt{1-K^2}},for 0\le K\le 1\backslash tan(\theta \theta )\; =\; \backslash frac\{K\}\{\backslash sqrt\{1\; –\; K^2\}\},\; \backslash quad\; \backslash text\{for\; \}\; 0\; \backslash le\; K\; \backslash le\; 1$tan(θθ)=1−K2​K​,for 0≤K≤1

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