Subjective Type

Solve the equation for $$x : {\sin}^{-1}\dfrac{5}{x} + {\sin}^{-1}\dfrac{12}{x}=\dfrac{\pi}{2}, x \ne 0$$.

Solution

The given equation is
$${\sin}^{-1}\dfrac{5}{x}+{\sin}^{-1}\dfrac{12}{x}=\dfrac{\pi}{2}$$

$$\Rightarrow \dfrac{12}{x}=\sin \begin{pmatrix}\dfrac{\pi}{2}-{\sin}^{-1}\dfrac{5}{x}\end{pmatrix}$$

$$\Rightarrow \dfrac{12}{x}=\cos \begin{pmatrix}{\sin}^{-1}\dfrac{5}{x}\end{pmatrix}$$

$$\Rightarrow \dfrac{12}{x}=\sqrt{1-\begin{pmatrix}\dfrac{5}{x}\end{pmatrix}^2}$$ $$\because {\cos}({\sin}^{-1}x)=\sqrt{1-x^2}$$

$$\begin{pmatrix}\dfrac{12}{x}\end{pmatrix}^2 = 1-\dfrac{25}{x^2}$$

$$\Rightarrow \dfrac{144}{x^2}+\dfrac{25}{x^2}=1$$


$$\Rightarrow \dfrac{169}{x^2}=1$$

$$x^2=169$$

$$\Rightarrow x = \pm \sqrt{169}$$

$$\Rightarrow x=13, -13$$

But $$x=-13$$ doesn't satisfy the given equation.
Hence the solution of the given equation is $$x=13$$.

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