Verified Answer :

looking at that picture
we could use the pythagorean theorem to find the adjacent side,
or cosine value
\(\bf tan\left[ sin^{-1}(x) \right]\implies \begin{cases} x=sin(\theta)\\ or\\ \cfrac{x}{1}=sin(\theta)\implies \cfrac{x=opposite=b}{1=radius=c} \end{cases}\\ \\ \quad \\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \\ \quad \\ \sqrt{1^2-x^2}=a=cos(\theta)\\ ----------------------------\\ thus \\ \quad \\ tan\left[ sin^{-1}(x) \right]\implies tan(\theta)\implies \cfrac{sin(\theta)}{cos(\theta)}\implies \cfrac{\frac{x}{1}}{\sqrt{1-x^2}} \\ \quad \\ \cfrac{x}{\sqrt{1-x^2}}\)

View image jdoe0001
lukyo

If you're using the app, try seeing this answer through your browser:  https://brainly.com/question/2799412

_______________


Let  \(\mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}\)

(that is the range of the inverse sine function).


So,

\(\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}\)


Square both sides:

\(\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}\)


Since \(\mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},}\) then \(\mathsf{cos\,\theta}\) is positive. So take the positive square root and you get

\(\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}\)


Then,

\(\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}\)


I hope this helps. =)


Tags:  inverse trigonometric function sin tan arcsin trigonometry

Related Questions