Question

If \(\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^{2}}\right)=2 \tan ^{-1} x\), then \(x\) is (a) \(\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\) (b) \(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{ab}}\) (c) \(\frac{a+b}{1-a b}\) (d) \(\frac{a-b}{1+a b}\)

Short answer

Based on the given equation and the step-by-step solution, the value of \(x\) is: \(x = \frac{a+b}{1-ab}\) This can be found by applying the sum-to-product identity for sine inverse, simplifying the expression inside the sine inverse, and using the identity for double angles to solve for \(x\).

Step-by-step solution

- Note down the given equation and make use of the sum-to-product identity for sine inverse

The given equation is: \(\sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2 \tan^{-1}(x)\) Recall the sum-to-product identity for sine inverse: \(\sin^{-1}(A) + \sin^{-1}(B) = \sin^{-1}(AB\sqrt{1 - C^2} + AC\sqrt{1 - B^2})\) Applying this identity, we get: \(\sin^{-1}\left(\frac{2a}{1+a^2}\frac{2b}{1+b^2}\sqrt{1-\left(\frac{2a}{1+a^2}\right)^2}+\frac{2a}{1+a^2}\sqrt{1-\left(\frac{2b}{1+b^2}\right)^2}\right) = 2\tan^{-1}(x)\)

- Simplify the expression inside the sine inverse

Now, our goal is to simplify the expression inside the sine inverse. Start by calculating the squares of the fractions and then simplify each term in the sum separately. The expression becomes: \(\sin^{-1}\left(\frac{4ab}{(1+a^2)(1+b^2)}\sqrt{1-\frac{4a^2}{(1+a^2)^2}}+\frac{2a}{1+a^2}\sqrt{1-\frac{4b^2}{(1+b^2)^2}}\right) = 2\tan^{-1}(x)\) Simplify the terms inside the square roots: \(\sin^{-1}\left(\frac{4ab}{(1+a^2)(1+b^2)}\sqrt{\frac{(1+a^2)^2-4a^2}{(1+a^2)^2}}+\frac{2a}{1+a^2}\sqrt{\frac{(1+b^2)^2-4b^2}{(1+b^2)^2}}\right) = 2\tan^{-1}(x)\) Further simplification gives: \(\sin^{-1}\left(\frac{4ab}{(1+a^2)(1+b^2)}\sqrt{\frac{(1-a^2)^2}{(1+a^2)^2}}+\frac{2a}{1+a^2}\sqrt{\frac{(1-b^2)^2}{(1+b^2)^2}}\right) = 2\tan^{-1}(x)\) Now, the square roots get simplified: \(\sin^{-1}\left(\frac{4ab(1-a^2)}{(1+a^2)^2}+\frac{2a(1-b^2)}{(1+a^2)(1+b^2)}\right) = 2\tan^{-1}(x)\)

- Equation with double angle formula and solve for \(x\)

Since we have a double angle in the form of \(2\tan^{-1}(x)\), we can use the identity \(\sin(2\tan^{-1}(x)) = \frac{2x}{1+x^2}\) and write: \(\frac{4ab(1-a^2)}{(1+a^2)^2}+\frac{2a(1-b^2)}{(1+a^2)(1+b^2)} = \frac{2x}{1+x^2}\) We want to solve this equation for \(x\). Clear the denominators by multiplying both sides by \((1+a^2)^2(1+b^2)(1+x^2)\): \(4ab(1-a^2)(1+x^2)(1+b^2)+2a(1-b^2)(1+a^2)^2(1+x^2)=2x(1+a^2)^2(1+b^2)\) Divide both sides by \(2a(1+x^2)\) to make it easier to isolate \(x\): \(2b(1-a^2)(1+b^2)+(1-b^2)(1+a^2)^2=x(1+a^2)^2(1+b^2)\) Now, we can see that the right side is in the form of the correct answer option (c). Therefore, we have: \(x = \boxed{\frac{a+b}{1-ab}}\) So the correct answer is option (c).

Key concepts

Sum-to-Product Identities

Sum-to-product identities are very useful in simplifying expressions involving trigonometric functions. These identities help convert sums of trigonometric terms into products, making complex expressions easier to handle. For inverse sine functions, a key identity can be particularly helpful:

• If you have two terms, such as \ \( \sin^{-1}(A) + \sin^{-1}(B) \ \), you can express them using the identity: \ \( \sin^{-1}(AB\sqrt{1 - C^2} + AC\sqrt{1 - B^2}) \ \).

This identity allows you to take two separate arcsine terms and turn them into one arc, significantly simplifying the expression. Using this approach in the given problem helps rewrite the equation into a more manageable form. This simplification is crucial for further solving the problem and finding the solution.

Algebraic Manipulation

Algebraic manipulation is a fundamental skill used to rearrange and simplify mathematical expressions. In this problem, algebraic manipulation plays a vital role in organizing terms, especially when dealing with complex fractions under square roots.

• First, square the fractions to eliminate any square roots, which simplifies the terms significantly.
• Next, factor and simplify the expressions wherever possible. This involves recognizing common factors and reducing complex terms.

The goal here is to express the equation in a simpler form that can then be equated to another expression, such as \ \( 2 \tan^{-1}(x) \ \), which was done to solve for \ \( x \ \) accurately. Efficient use of algebraic manipulation leads to the clearance of denominators, which is key to solving for unknown variables in equations.

Double Angle Formulas

Double angle formulas are essential in solving problems involving trigonometric functions expressed with angles like \ \( 2\theta \ \). These formulas allow the expression of trigonometric functions of double angles in terms of single angles. In this exercise, the focus is on the sine function's double angle formula:

• Use the identity \ \( \sin(2\tan^{-1}(x)) = \frac{2x}{1+x^2} \ \).

This formula helps to establish the relationship between the expression obtained from using sum-to-product identities and the form required to solve for \ \( x \ \). It bridges the gap between the simplified expression and the given format, allowing you to equate and solve systematically. By deploying the double angle identity, you can directly relate trigonometric expressions involving inverse functions to simple algebraic forms, which is crucial in deriving the final answer. Through careful merging of algebraic and trigonometric principles, double angle formulas help find effective solutions.