Question

If sin (cot$^{-1}$ (cos (tan$^{-1}$x))) = $\sqrt{\frac{P}{18}}$ when x = 4 then the value of $P$ is

Answer 1

17

Answer 2

To find the value of $P$, we need to evaluate the given expression step-by-step from the innermost function outwards, substituting $x=4$.

The given equation is: $\sin (\cot^{-1} (\cos (\tan^{-1}x))) = \sqrt{\frac{P}{18}}$

Substitute $x=4$ into the equation: $\sin (\cot^{-1} (\cos (\tan^{-1}4))) = \sqrt{\frac{P}{18}}$

Step 1: Evaluate the innermost expression, $\tan^{-1}4$. Let $A = \tan^{-1}4$. This implies $\tan A = 4$. We can construct a right-angled triangle where one angle is $A$. Since $\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{1}$, we have: Opposite side = 4 Adjacent side = 1 Using the Pythagorean theorem, the hypotenuse = $\sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{4^2 + 1^2} = \sqrt{16+1} = \sqrt{17}$.

Step 2: Evaluate $\cos(\tan^{-1}4)$, which is $\cos A$. From the triangle constructed in Step 1: $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{17}}$.

Step 3: Evaluate $\cot^{-1}(\cos(\tan^{-1}4))$, which is $\cot^{-1}\left(\frac{1}{\sqrt{17}}\right)$. Let $B = \cot^{-1}\left(\frac{1}{\sqrt{17}}\right)$. This implies $\cot B = \frac{1}{\sqrt{17}}$. We can construct another right-angled triangle where one angle is $B$. Since $\cot B = \frac{\text{adjacent}}{\text{opposite}} = \frac{1}{\sqrt{17}}$, we have: Adjacent side = 1 Opposite side = $\sqrt{17}$ Using the Pythagorean theorem, the hypotenuse = $\sqrt{(\text{adjacent})^2 + (\text{opposite})^2} = \sqrt{1^2 + (\sqrt{17})^2} = \sqrt{1+17} = \sqrt{18}$.

Step 4: Evaluate $\sin(\cot^{-1}(\cos(\tan^{-1}4)))$, which is $\sin B$. From the triangle constructed in Step 3: $\sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{17}}{\sqrt{18}}$.

Step 5: Equate the result to $\sqrt{\frac{P}{18}}$ and solve for $P$. We have $\frac{\sqrt{17}}{\sqrt{18}} = \sqrt{\frac{P}{18}}$. To eliminate the square roots, square both sides of the equation: $\left(\frac{\sqrt{17}}{\sqrt{18}}\right)^2 = \left(\sqrt{\frac{P}{18}}\right)^2$ $\frac{17}{18} = \frac{P}{18}$ Multiply both sides by 18: $P = 17$.