Introduction

Similar to standard trigonometric functions, inverse trigonometric functions have powerful formulas for the sum and difference of functions. These are essential for simplifying complex expressions and solving equations. The key to understanding them is to see how they derive from the standard sum/difference formulas (e.g., tan(A+B)).

1. Tangent Sum and Difference Formulas ➕➖

These are the most commonly used and important formulas in this category.

  • tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy),if xy<1\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right), \quad \text{if } xy < 1
  • tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy),if x>0,y>0, and xy>1\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right), \quad \text{if } x>0, y>0, \text{ and } xy > 1
  • tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy),if xy>−1\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right), \quad \text{if } xy > -1

Derivation/Intuition: These formulas come directly from the standard tan(A+B) identity. Let A=tan⁡−1xA=\tan^{-1}x and B=tan⁡−1yB=\tan^{-1}y. This means tan⁡A=x\tan A = x and tan⁡B=y\tan B = y.

We know that tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B=x+y1−xy\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B} = \frac{x+y}{1-xy}

Taking the tan⁻¹ of both sides gives A+B=tan⁡−1(x+y1−xy)A+B = \tan^{-1}\left(\frac{x+y}{1-xy}\right), which proves the basic formula.

The additional π in the second formula is needed when the sum of the angles A+BA+B exceeds π/2\pi/2, falling outside the principal value branch of tan⁻¹.

Example 1: Find the value of tan⁡−1(1/2)+tan⁡−1(1/3)\tan^{-1}(1/2) + \tan^{-1}(1/3).

  • Here x=1/2,y=1/3x=1/2, y=1/3. xy=1/6<1xy = 1/6 < 1, so we use the first formula.

  • Value = tan⁡−1(1/2+1/31−(1/2)(1/3))=tan⁡−1(5/65/6)=tan⁡−1(1)=π/4\tan^{-1}\left(\frac{1/2+1/3}{1-(1/2)(1/3)}\right) = \tan^{-1}\left(\frac{5/6}{5/6}\right) = \tan^{-1}(1) = \pi/4.

Example 2: Find the value of tan⁡−1(2)+tan⁡−1(3)\tan^{-1}(2) + \tan^{-1}(3).

  • Here x=2,y=3x=2, y=3. x>0,y>0x>0, y>0 and xy=6>1xy = 6 > 1, so we use the second formula.

  • Value = π+tan⁡−1(2+31−2⋅3)=π+tan⁡−1(5−5)=π+tan⁡−1(−1)=π−π/4=3π/4\pi + \tan^{-1}\left(\frac{2+3}{1-2 \cdot 3}\right) = \pi + \tan^{-1}\left(\frac{5}{-5}\right) = \pi + \tan^{-1}(-1) = \pi - \pi/4 = 3\pi/4.

2. Sine Sum and Difference Formulas

These are derived from the sin(A±B) formulas. They are less commonly used than the tangent formulas because of their complex conditions for validity.

  • sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2)\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})
  • sin⁡−1x−sin⁡−1y=sin⁡−1(x1−y2−y1−x2)\sin^{-1}x - \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} - y\sqrt{1-x^2})

Derivation/Intuition: Let A=sin⁡−1xA=\sin^{-1}x and B=sin⁡−1yB=\sin^{-1}y. Then sin⁡A=x\sin A = x and sin⁡B=y\sin B = y.

Using right triangles, we find cos⁡A=1−x2\cos A = \sqrt{1-x^2} and cos⁡B=1−y2\cos B = \sqrt{1-y^2}.

The formula for sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B becomes x1−y2+y1−x2x\sqrt{1-y^2} + y\sqrt{1-x^2}. Taking sin⁻¹ of both sides gives the identity.

Example: Find the value of sin⁡−1(3/5)+sin⁡−1(8/17)\sin^{-1}(3/5) + \sin^{-1}(8/17).

  • x=3/5,y=8/17x=3/5, y=8/17. The formula gives:

  • sin⁡−1(351−(817)2+8171−(35)2)\sin^{-1}\left(\frac{3}{5}\sqrt{1-(\frac{8}{17})^2} + \frac{8}{17}\sqrt{1-(\frac{3}{5})^2}\right)

  • =sin⁡−1(35225289+8171625)=sin⁡−1(35⋅1517+817⋅45)= \sin^{-1}\left(\frac{3}{5}\sqrt{\frac{225}{289}} + \frac{8}{17}\sqrt{\frac{16}{25}}\right) = \sin^{-1}\left(\frac{3}{5} \cdot \frac{15}{17} + \frac{8}{17} \cdot \frac{4}{5}\right)

  • =sin⁡−1(4585+3285)=sin⁡−1(7785)= \sin^{-1}\left(\frac{45}{85} + \frac{32}{85}\right) = \sin^{-1}\left(\frac{77}{85}\right)

3. Cosine Sum and Difference Formulas

These are derived from the cos(A±B) formulas and also have complex conditions for general use.

  • cos⁡−1x+cos⁡−1y=cos⁡−1(xy−1−x21−y2)\cos^{-1}x + \cos^{-1}y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})
  • cos⁡−1x−cos⁡−1y=cos⁡−1(xy+1−x21−y2)\cos^{-1}x - \cos^{-1}y = \cos^{-1}(xy + \sqrt{1-x^2}\sqrt{1-y^2})

Example: Find the value of cos⁡−1(4/5)+cos⁡−1(12/13)\cos^{-1}(4/5) + \cos^{-1}(12/13).

  • x=4/5,y=12/13x=4/5, y=12/13. The formula gives:
  • cos⁡−1(45⋅1213−1−(45)21−(1213)2)\cos^{-1}\left(\frac{4}{5} \cdot \frac{12}{13} - \sqrt{1-(\frac{4}{5})^2}\sqrt{1-(\frac{12}{13})^2}\right)
  • =cos⁡−1(4865−35⋅513)=cos⁡−1(4865−1565)=cos⁡−1(3365)= \cos^{-1}\left(\frac{48}{65} - \frac{3}{5} \cdot \frac{5}{13}\right) = \cos^{-1}\left(\frac{48}{65} - \frac{15}{65}\right) = \cos^{-1}\left(\frac{33}{65}\right).

4. Multiple Angle Formulas 📐

These are extremely useful formulas, especially for substitutions in calculus. They are all derived from the standard double angle formulas sin(2θ), cos(2θ), and tan(2θ) by letting x = tan(θ).

  • 2tan⁡−1x=tan⁡−1(2x1−x2),for ∣x∣<12\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right), \quad \text{for } |x| < 1
  • 2tan⁡−1x=sin⁡−1(2x1+x2),for ∣x∣≤12\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^2}\right), \quad \text{for } |x| \le 1
  • 2tan⁡−1x=cos⁡−1(1−x21+x2),for x≥02\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), \quad \text{for } x \ge 0

Example: Prove that 2tan⁡−1(1/3)=cos⁡−1(4/5)2\tan^{-1}(1/3) = \cos^{-1}(4/5).

  • Use the formula 2tan⁡−1x=cos⁡−1(1−x21+x2)2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) with x=1/3x=1/3.
  • 2tan⁡−1(1/3)=cos⁡−1(1−(1/3)21+(1/3)2)=cos⁡−1(1−1/91+1/9)2\tan^{-1}(1/3) = \cos^{-1}\left(\frac{1-(1/3)^2}{1+(1/3)^2}\right) = \cos^{-1}\left(\frac{1-1/9}{1+1/9}\right).
  • =cos⁡−1(8/910/9)=cos⁡−1(810)=cos⁡−1(4/5)= \cos^{-1}\left(\frac{8/9}{10/9}\right) = \cos^{-1}\left(\frac{8}{10}\right) = \cos^{-1}(4/5). The identity is proven.

Example 1: Sum of Tangent Inverses (xy < 1)

Question: Find the value of tan⁡−1(1/2)+tan⁡−1(1/3)\tan^{-1}(1/2) + \tan^{-1}(1/3).

Explanation:

  1. Check the Condition: Let x=1/2x=1/2 and y=1/3y=1/3. The product xy=(1/2)(1/3)=1/6xy = (1/2)(1/3) = 1/6.

    Since xy<1xy < 1, we can use the primary sum formula.

  2. Apply the Formula: tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) tan⁡−1(1/2)+tan⁡−1(1/3)\tan^{-1}(1/2) + \tan^{-1}(1/3) =tan⁡−1(1/2+1/31−(1/2)(1/3))= \tan^{-1}\left(\frac{1/2+1/3}{1-(1/2)(1/3)}\right) =tan⁡−1(5/61−1/6)= \tan^{-1}\left(\frac{5/6}{1-1/6}\right) =tan⁡−1(5/65/6)= \tan^{-1}\left(\frac{5/6}{5/6}\right) =tan⁡−1(1)=π4= \tan^{-1}(1) = \frac{\pi}{4}

Answer: π/4\pi/4.

Example 2: Sum of Tangent Inverses (xy > 1)

Question: Find the value of tan⁡−1(2)+tan⁡−1(3)\tan^{-1}(2) + \tan^{-1}(3).

Explanation:

  1. Check the Condition: Let x=2x=2 and y=3y=3. We have x>0,y>0x>0, y>0 and the product xy=6>1xy = 6 > 1.

    Therefore, we must use the formula that includes the π\pi term.

  2. Apply the Formula: tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) tan⁡−1(2)+tan⁡−1(3)\tan^{-1}(2) + \tan^{-1}(3) =π+tan⁡−1(2+31−2⋅3)= \pi + \tan^{-1}\left(\frac{2+3}{1-2 \cdot 3}\right) =π+tan⁡−1(5−5)= \pi + \tan^{-1}\left(\frac{5}{-5}\right) =π+tan⁡−1(−1)= \pi + \tan^{-1}(-1)

  3. Evaluate: Since tan⁡−1(−1)=−π/4\tan^{-1}(-1) = -\pi/4, the result is π−π/4=3π4\pi - \pi/4 = \frac{3\pi}{4}.

Answer: 3π/43\pi/4.

Example 3: Solving an Equation with tan⁡−1\tan^{-1}

Question: Solve the equation tan⁡−1(2x)+tan⁡−1(3x)=π/4\tan^{-1}(2x) + \tan^{-1}(3x) = \pi/4.

Explanation:

  1. Apply the Sum Formula: Assuming the condition (2x)(3x)<1(2x)(3x) < 1 holds, we apply the sum formula to the left side.

    tan⁡−1(2x+3x1−(2x)(3x))=tan⁡−1(5x1−6x2)=π/4\tan^{-1}\left(\frac{2x+3x}{1-(2x)(3x)}\right) = \tan^{-1}\left(\frac{5x}{1-6x^2}\right) = \pi/4.

  2. Solve the Equation: Take the tangent of both sides: 5x1−6x2=tan⁡(π/4)=1\frac{5x}{1-6x^2} = \tan(\pi/4) = 1 5x=1−6x2  ⟹  6x2+5x−1=05x = 1-6x^2 \implies 6x^2+5x-1=0

  3. Find Potential Solutions: Factor the quadratic: (6x−1)(x+1)=0(6x-1)(x+1)=0.

    The potential solutions are x=1/6x=1/6 and x=−1x=-1.

  4. Check the Initial Condition: We must check if these solutions satisfy our assumption 6x2<16x^2 < 1.

    • For x=1/6x=1/6: 6(1/6)2=6/36=1/6<16(1/6)^2 = 6/36 = 1/6 < 1. This solution is valid.

    • For x=−1x=-1: 6(−1)2=6>16(-1)^2 = 6 > 1. This solution is invalid.

Answer: x=1/6x=1/6.

Example 4: Sum of Sine Inverses

Question: Find the value of sin⁡−1(3/5)+sin⁡−1(8/17)\sin^{-1}(3/5) + \sin^{-1}(8/17).

Explanation:

  1. State the Formula: sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2)\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})

  2. Calculate Intermediate Values: Let x=3/5x=3/5 and y=8/17y=8/17. These are ratios from Pythagorean triplets (3-4-5 and 8-15-17).

    • 1−x2=1−(3/5)2=16/25=4/5\sqrt{1-x^2} = \sqrt{1-(3/5)^2} = \sqrt{16/25} = 4/5.

    • 1−y2=1−(8/17)2=225/289=15/17\sqrt{1-y^2} = \sqrt{1-(8/17)^2} = \sqrt{225/289} = 15/17.

  3. Substitute and Solve: sin⁡−1(35⋅1517+817⋅45)\sin^{-1}\left(\frac{3}{5} \cdot \frac{15}{17} + \frac{8}{17} \cdot \frac{4}{5}\right) =sin⁡−1(4585+3285)= \sin^{-1}\left(\frac{45}{85} + \frac{32}{85}\right) =sin⁡−1(7785)= \sin^{-1}\left(\frac{77}{85}\right)

Answer: sin⁡−1(77/85)\sin^{-1}(77/85).

Example 5: Using 2tan⁡−1x2\tan^{-1}x Formula

Question: Show that 2tan⁡−1(1/3)+tan⁡−1(1/7)=π/42\tan^{-1}(1/3) + \tan^{-1}(1/7) = \pi/4.

Explanation: Part 1: Simplify the 2tan⁡−12\tan^{-1} term.

  1. Use the formula 2tan⁡−1x=tan⁡−1(2x1−x2)2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) Here x=1/3x=1/3.
  2. 2tan⁡−1(1/3)=tan⁡−1(2(1/3)1−(1/3)2)2\tan^{-1}(1/3) = \tan^{-1}\left(\frac{2(1/3)}{1-(1/3)^2}\right) =tan⁡−1(2/31−1/9)= \tan^{-1}\left(\frac{2/3}{1-1/9}\right) =tan⁡−1(2/38/9)= \tan^{-1}\left(\frac{2/3}{8/9}\right) =tan⁡−1(23×98)= \tan^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) =tan⁡−1(3/4)= \tan^{-1}(3/4).

Part 2: Solve the new sum. The problem is now to evaluate tan⁡−1(3/4)+tan⁡−1(1/7)\tan^{-1}(3/4) + \tan^{-1}(1/7).

  1. Check the condition: xy=(3/4)(1/7)=3/28<1xy=(3/4)(1/7)=3/28 < 1. We use the standard sum formula.
  2. tan⁡−1(3/4+1/71−(3/4)(1/7))\tan^{-1}\left(\frac{3/4+1/7}{1-(3/4)(1/7)}\right) =tan⁡−1((21+4)/281−3/28)= \tan^{-1}\left(\frac{(21+4)/28}{1-3/28}\right) =tan⁡−1(25/2825/28)=tan⁡−1(1)=π/4= \tan^{-1}\left(\frac{25/28}{25/28}\right) = \tan^{-1}(1) = \pi/4.

Answer: Proven.

Example 6: Sum of Three Terms

Question: Find the value of tan⁡−1(1)+tan⁡−1(2)+tan⁡−1(3)\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3).

Explanation: We add the terms in pairs.

  1. Sum tan⁡−1(2)+tan⁡−1(3)\tan^{-1}(2) + \tan^{-1}(3):

    Here x=2,y=3x=2, y=3. Since x>0,y>0x>0, y>0 and xy=6>1xy=6>1, we use the formula with π\pi.

    tan⁡−1(2)+tan⁡−1(3)\tan^{-1}(2) + \tan^{-1}(3) =π+tan⁡−1(2+31−6)= \pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) =π+tan⁡−1(−1)= \pi + \tan^{-1}(-1) =π−π/4=3π/4= \pi - \pi/4 = 3\pi/4

  2. Add the remaining term: The total sum is now tan⁡−1(1)+3π/4\tan^{-1}(1) + 3\pi/4.

  3. Evaluate: We know tan⁡−1(1)=π/4\tan^{-1}(1) = \pi/4. So the sum is π/4+3π/4=π\pi/4 + 3\pi/4 = \pi.

Answer: π\pi.

Example 7: Using 2tan⁡−1x=cos⁡−1(...)2\tan^{-1}x = \cos^{-1}(...)

Question: Find the value of 2tan⁡−1(1/2)2\tan^{-1}(1/2).

Explanation: We can express the result in terms of a different inverse function. One of the most useful identities is: 2tan⁡−1x=cos⁡−1(1−x21+x2)2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)

  1. Apply the Formula: Substitute x=1/2x=1/2.

    2tan⁡−1(1/2)=cos⁡−1(1−(1/2)21+(1/2)2)2\tan^{-1}(1/2) = \cos^{-1}\left(\frac{1-(1/2)^2}{1+(1/2)^2}\right).

  2. Simplify the argument: cos⁡−1(1−1/41+1/4)\cos^{-1}\left(\frac{1-1/4}{1+1/4}\right) =cos⁡−1(3/45/4)=cos⁡−1(3/5)= \cos^{-1}\left(\frac{3/4}{5/4}\right) = \cos^{-1}(3/5).

Answer: The value can be expressed as cos⁡−1(3/5)\cos^{-1}(3/5).

Example 8: Converting to Tangent Inverses

Question: Find the value of sin⁡−1(4/5)+2tan⁡−1(1/3)\sin^{-1}(4/5) + 2\tan^{-1}(1/3).

Explanation:

  1. Strategy: The easiest way to solve this is to have all terms in the tan⁻¹ form.

  2. Convert sin⁡−1(4/5)\sin^{-1}(4/5): Let α=sin⁡−1(4/5)\alpha = \sin^{-1}(4/5). This corresponds to a 3-4-5 right triangle. So, tan⁡α=4/3\tan\alpha = 4/3. Thus, sin⁡−1(4/5)=tan⁡−1(4/3)\sin^{-1}(4/5) = \tan^{-1}(4/3).

  3. Simplify 2tan⁡−1(1/3)2\tan^{-1}(1/3): From Example 5, we know 2tan⁡−1(1/3)=tan⁡−1(3/4)2\tan^{-1}(1/3) = \tan^{-1}(3/4).

  4. Calculate the Sum: The expression is now tan⁡−1(4/3)+tan⁡−1(3/4)\tan^{-1}(4/3) + \tan^{-1}(3/4)

  5. Use the Cofunction Identity: We use the identity tan⁡−1(z)+cot⁡−1(z)=π/2\tan^{-1}(z) + \cot^{-1}(z) = \pi/2

    We can rewrite tan⁡−1(3/4)\tan^{-1}(3/4) as cot⁡−1(4/3)\cot^{-1}(4/3).

    The sum becomes tan⁡−1(4/3)+cot⁡−1(4/3)\tan^{-1}(4/3) + \cot^{-1}(4/3), which is equal to π/2\pi/2.

Answer: π/2\pi/2.

Example 9: Solving a More Complex Equation

Question: Solve for x: sin⁡−1(1−x)−2sin⁡−1x=π/2\sin^{-1}(1-x) - 2\sin^{-1}x = \pi/2.

Explanation:

  1. Rearrange and Apply Sine: Rearrange the equation to sin⁡−1(1−x)=π/2+2sin⁡−1x\sin^{-1}(1-x) = \pi/2 + 2\sin^{-1}x

    Now, take the sine of both sides. 1−x=sin⁡(π/2+2sin⁡−1x)1-x = \sin(\pi/2 + 2\sin^{-1}x)

  2. Use Trigonometric Identities: Use the identity sin⁡(π/2+θ)=cos⁡(θ)\sin(\pi/2 + \theta) = \cos(\theta). 1−x=cos⁡(2sin⁡−1x)1-x = \cos(2\sin^{-1}x) Now use the identity cos⁡(2A)=1−2sin⁡2A\cos(2A) = 1-2\sin^2A. Let A=sin⁡−1xA = \sin^{-1}x. 1−x=1−2sin⁡2(sin⁡−1x)=1−2(sin⁡(sin⁡−1x))21-x = 1 - 2\sin^2(\sin^{-1}x) = 1 - 2(\sin(\sin^{-1}x))^2

  3. Simplify and Solve:

    1−x=1−2x21-x = 1-2x^2.

    This simplifies to 2x2−x=0  ⟹  x(2x−1)=02x^2-x=0 \implies x(2x-1)=0.

    The potential solutions are x=0x=0 and x=1/2x=1/2.

  4. Check for Validity: We must substitute the potential solutions back into the original equation.

    • Check x=0: sin⁡−1(1)−2sin⁡−1(0)=π/2−2(0)=π/2\sin^{-1}(1) - 2\sin^{-1}(0) = \pi/2 - 2(0) = \pi/2. This is a valid solution.

    • Check x=1/2: sin⁡−1(1/2)−2sin⁡−1(1/2)=−sin⁡−1(1/2)=−π/6\sin^{-1}(1/2) - 2\sin^{-1}(1/2) = -\sin^{-1}(1/2) = -\pi/6. This is not equal to π/2\pi/2. This is not a valid solution.

Answer: x=0x=0.

Example 10: Sum of Cosine Inverses

Question: Find the value of cos⁡−1(4/5)+cos⁡−1(12/13)\cos^{-1}(4/5) + \cos^{-1}(12/13).

Explanation:

  1. State the Formula: cos⁡−1x+cos⁡−1y=cos⁡−1(xy−1−x21−y2)\cos^{-1}x + \cos^{-1}y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})

  2. Identify Pythagorean Triplets: Let x=4/5x=4/5 and y=12/13y=12/13. These are from 3-4-5 and 5-12-13 triangles.

    • 1−x2=1−(4/5)2=3/5\sqrt{1-x^2} = \sqrt{1-(4/5)^2} = 3/5.

    • 1−y2=1−(12/13)2=5/13\sqrt{1-y^2} = \sqrt{1-(12/13)^2} = 5/13.

  3. Substitute and Solve: cos⁡−1(45⋅1213−35⋅513)\cos^{-1}\left(\frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13}\right) =cos⁡−1(4865−1565)=cos⁡−1(3365)= \cos^{-1}\left(\frac{48}{65} - \frac{15}{65}\right) = \cos^{-1}\left(\frac{33}{65}\right)

Answer: cos⁡−1(33/65)\cos^{-1}(33/65).

Example 11: Solving an equation with 2tan⁡−1x2\tan^{-1}x

Question: Solve for x: 2tan⁡−1(sin⁡x)=tan⁡−1(2sec⁡x)2\tan^{-1}(\sin x) = \tan^{-1}(2\sec x), for x∈(0,π/2)x \in (0, \pi/2).

Explanation:

  1. Apply the 2tan⁡−1u2\tan^{-1}u Identity: Let u=sin⁡xu = \sin x. The LHS becomes tan⁡−1(2sin⁡x1−sin⁡2x)=tan⁡−1(2sin⁡xcos⁡2x)\tan^{-1}\left(\frac{2\sin x}{1-\sin^2x}\right) = \tan^{-1}\left(\frac{2\sin x}{\cos^2x}\right)

  2. Equate Arguments: The equation is now tan⁡−1(2sin⁡xcos⁡2x)=tan⁡−1(2sec⁡x)\tan^{-1}(\frac{2\sin x}{\cos^2x}) = \tan^{-1}(2\sec x) This implies: 2sin⁡xcos⁡2x=2sec⁡x=2cos⁡x\frac{2\sin x}{\cos^2x} = 2\sec x = \frac{2}{\cos x}

  3. Solve for x: Since x∈(0,π/2)x \in (0, \pi/2), cos⁡x≠0\cos x \ne 0, so we can multiply both sides by cos⁡2x/2\cos^2x / 2. sin⁡x=cos⁡x\sin x = \cos x. This simplifies to tan⁡x=1\tan x = 1.

  4. Find the Solution: The only solution for tan⁡x=1\tan x=1 in the interval (0,π/2)(0, \pi/2) is x=π/4x=\pi/4.

Answer: x=π/4x=\pi/4.