Math Problem Statement

tan(sin^(-1)x)

Solution

To find tan(sin1(x))\tan(\sin^{-1}(x)), we can proceed with the following steps:

Step-by-step solution:

  1. Let θ=sin1(x)\theta = \sin^{-1}(x), so that sin(θ)=x\sin(\theta) = x.
  2. Now, we need to find tan(θ)\tan(\theta). Using the identity for the tangent in terms of sine and cosine: tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
  3. Since sin(θ)=x\sin(\theta) = x, we now need to find cos(θ)\cos(\theta). Use the Pythagorean identity for sine and cosine: sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 Substituting sin(θ)=x\sin(\theta) = x: x2+cos2(θ)=1x^2 + \cos^2(\theta) = 1 Solving for cos(θ)\cos(\theta): cos(θ)=1x2\cos(\theta) = \sqrt{1 - x^2}
  4. Now, substitute the values of sin(θ)\sin(\theta) and cos(θ)\cos(\theta) into the tangent expression: tan(θ)=x1x2\tan(\theta) = \frac{x}{\sqrt{1 - x^2}}

Thus, tan(sin1(x))=x1x2\tan(\sin^{-1}(x)) = \frac{x}{\sqrt{1 - x^2}}.

Final answer:

tan(sin1(x))=x1x2\tan(\sin^{-1}(x)) = \frac{x}{\sqrt{1 - x^2}}

Do you need further details or have any questions?

Here are 5 related questions for further exploration:

  1. What is tan(cos1(x))\tan(\cos^{-1}(x))?
  2. How would you find tan(sin1(x)+cos1(x))\tan(\sin^{-1}(x) + \cos^{-1}(x))?
  3. What is the range of values for which tan(sin1(x))\tan(\sin^{-1}(x)) is valid?
  4. How does the identity change for inverse trigonometric functions of complex numbers?
  5. Can you derive tan(sin1(2x))\tan(\sin^{-1}(2x))?

Tip: Always check the domain of the function when dealing with inverse trigonometric functions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Inverse Trigonometric Functions
Tangent
Pythagorean Identity

Formulas

tan(θ) = sin(θ) / cos(θ)
sin²(θ) + cos²(θ) = 1
tan(sin^(-1)(x)) = x / sqrt(1 - x²)

Theorems

Pythagorean Identity
Tangent Definition in Terms of Sine and Cosine

Suitable Grade Level

Grades 10-12