Question

If
${cos}^{ - 1}x - {sin}^{ - 1}x = 0$

Find the value of x.​

Answer 1

Answer:

1/√2

Step-by-step explanation:

                     Method 1:

Given, cos⁻¹x - sin⁻¹ = 0       ...(1)

 We know cos⁻¹θ + sin⁻¹θ = π/2  is defined for all possible values.

 ∴ cos⁻¹x + sin⁻¹x = π/2         ...(2)

On adding (1) and (2), we get

   ⇒ 2cos⁻¹x = π/2        ⇒ cos⁻¹x = π/4

             ∴ cos(π/4) = x

             ⇒ 1/√2 = x

                    Method 2:

Given, cos⁻¹x - sin⁻¹x = 0

        ⇒ cos⁻¹x = sin⁻¹x

        ⇒ sin(cos⁻¹x) = sin(sin⁻¹x)

        ⇒ √(1 - x²) = x

        ⇒ 1 - x² = x²

        ⇒ 1 = 2x²

        ⇒ 1/√2 = x

Answer 2

$\large\underline{\sf{Solution-}}$

Given inverse Trigonometric equation is

$\rm :\longmapsto\: {cos}^{ - 1}x - {sin}^{ - 1}x = 0 - - - (1)$

We know,

$\rm :\longmapsto\: {cos}^{ - 1}x + {sin}^{ - 1}x = \dfrac{\pi}{2} - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\: {cos}^{ - 1}x + {sin}^{ - 1}x + {cos}^{ - 1}x - {sin}^{ - 1} x = \dfrac{\pi}{2}$

$\rm :\longmapsto\: 2{cos}^{ - 1}x = \dfrac{\pi}{2}$

$\rm :\longmapsto\: {cos}^{ - 1}x = \dfrac{\pi}{4}$

$\rm :\longmapsto\: x = cos\dfrac{\pi}{4}$

$\bf\implies \:x = \dfrac{1}{ \sqrt{2} }$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf & \bf \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2} & \sf x \: \in \: [ - 1, \: 1] \\ \\ \sf {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2} & \sf x \: \in \: ( - \infty , \: \infty ) \\ \\ \sf {sec}^{ - 1}x + {cosec}^{ - 1}x = \dfrac{\pi}{2} & \sf x \: \in \: ( - \infty ,1] \: \cup \: [1, \: \infty )\end{array}} \\ \end{gathered}$