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Question: If $\sin^{-1}\left(\frac{x}{\sqrt{x^2+5^2}}\right)=\tan^{-1}\left(\frac{x}{k}\right)$, then $k$ is e...

If sin⁡−1(xx2+52)=tan⁡−1(xk)\sin^{-1}\left(\frac{x}{\sqrt{x^2+5^2}}\right)=\tan^{-1}\left(\frac{x}{k}\right), then kk is equal to

Answer

5

Explanation

Solution

To solve the given equation sin⁡−1(xx2+52)=tan⁡−1(xk)\sin^{-1}\left(\frac{x}{\sqrt{x^2+5^2}}\right)=\tan^{-1}\left(\frac{x}{k}\right), we need to convert the left-hand side into a tan⁡−1\tan^{-1} form.

Let θ=sin⁡−1(xx2+52)\theta = \sin^{-1}\left(\frac{x}{\sqrt{x^2+5^2}}\right). This implies sin⁡θ=xx2+52\sin \theta = \frac{x}{\sqrt{x^2+5^2}}. The range of θ\theta is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

We can find cos⁡θ\cos \theta using the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta. cos⁡2θ=1−(xx2+52)2\cos^2 \theta = 1 - \left(\frac{x}{\sqrt{x^2+5^2}}\right)^2 cos⁡2θ=1−x2x2+52\cos^2 \theta = 1 - \frac{x^2}{x^2+5^2} cos⁡2θ=x2+52−x2x2+52\cos^2 \theta = \frac{x^2+5^2-x^2}{x^2+5^2} cos⁡2θ=52x2+52\cos^2 \theta = \frac{5^2}{x^2+5^2} Taking the square root, cos⁡θ=±5x2+52\cos \theta = \pm \frac{5}{\sqrt{x^2+5^2}}. Since θ∈[−π2,π2]\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}], cos⁡θ\cos \theta must be non-negative. Therefore, cos⁡θ=5x2+52\cos \theta = \frac{5}{\sqrt{x^2+5^2}}.

Now we can find tan⁡θ\tan \theta: tan⁡θ=sin⁡θcos⁡θ=xx2+525x2+52=x5\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{x}{\sqrt{x^2+5^2}}}{\frac{5}{\sqrt{x^2+5^2}}} = \frac{x}{5}.

So, θ=tan⁡−1(x5)\theta = \tan^{-1}\left(\frac{x}{5}\right). Note that θ\theta cannot be ±π2\pm \frac{\pi}{2} because sin⁡θ=±1\sin \theta = \pm 1 would imply x=±x2+52x = \pm \sqrt{x^2+5^2}, which leads to x2=x2+25x^2 = x^2+25, i.e., 0=250=25, which is impossible. Thus, θ∈(−π2,π2)\theta \in (-\frac{\pi}{2}, \frac{\pi}{2}), which is the principal range of tan⁡−1\tan^{-1}.

Substituting this back into the original equation: tan⁡−1(x5)=tan⁡−1(xk)\tan^{-1}\left(\frac{x}{5}\right) = \tan^{-1}\left(\frac{x}{k}\right).

For this equality to hold true for all valid xx, the arguments of the tan⁡−1\tan^{-1} functions must be equal: x5=xk\frac{x}{5} = \frac{x}{k}.

Assuming x≠0x \neq 0, we can cancel xx from both sides: 15=1k\frac{1}{5} = \frac{1}{k} k=5k = 5.

If x=0x=0, the equation becomes sin⁡−1(0)=tan⁡−1(0)\sin^{-1}(0) = \tan^{-1}(0), which is 0=00=0. This holds for any non-zero kk. However, the question asks for "k is equal to", implying a unique value that satisfies the equality for all relevant xx. Thus, k=5k=5 is the unique value.