Addition and Subtraction Formulas

Addition and Subtraction Formulas for Inverse Trigonometric Functions

Hello JEE aspirants! This lesson dives into the fascinating world of addition and subtraction formulas for inverse trigonometric functions. These formulas are indispensable tools for simplifying complex expressions and solving challenging JEE problems. Mastering them will significantly enhance your problem-solving speed and accuracy.

Think of inverse trigonometric functions as angles. Just as we have formulas to combine angles within sine, cosine, and tangent, we can also combine angles represented by $\sin^{-1}x$ , $\cos^{-1}x$ , and $\tan^{-1}x$ . The key is to understand the conditions under which these formulas hold true.

1. $\tan^{-1}x + \tan^{-1}y$ Formulas

Let's start with the tangent formulas. These are arguably the most frequently used in JEE problems.

$\tan^{-1}x + \tan^{-1}y = \begin{cases} \tan^{-1}\frac{x+y}{1-xy}, & \text{if } xy < 1 \\ \pi + \tan^{-1}\frac{x+y}{1-xy}, & \text{if } xy > 1, x > 0 \\ -\pi + \tan^{-1}\frac{x+y}{1-xy}, & \text{if } xy > 1, x < 0 \\ \pi/2, & \text{if } xy = 1, x > 0 \\ -\pi/2, & \text{if } xy = 1, x < 0 \end{cases}$

Derivation and Explanation:

Let $\tan^{-1}x = A$ and $\tan^{-1}y = B$ . Then, $x = \tan A$ and $y = \tan B$ . Now, consider $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{x+y}{1-xy}$ . Therefore, $A+B = \tan^{-1}\frac{x+y}{1-xy}$ .

The crucial part is understanding the condition $xy < 1$ . The range of $\tan^{-1}z$ is $(-\pi/2, \pi/2)$ . If $xy > 1$ , then $A+B$ falls outside this range. Let's look at an example.

If $x = \sqrt{3}$ and $y = \sqrt{3}$ , then $\tan^{-1}x = \frac{\pi}{3}$ and $\tan^{-1}y = \frac{\pi}{3}$ . So, $\tan^{-1}x + \tan^{-1}y = \frac{2\pi}{3}$ . But, $\tan^{-1}\frac{x+y}{1-xy} = \tan^{-1}\frac{2\sqrt{3}}{-2} = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$ . So, we need to add $\pi$ to get the correct value, $\frac{2\pi}{3}$ . In general, when $xy>1$ and $x>0$ , you need to add $\pi$ to the principal value.

Similarly, when $xy>1$ and $x<0$ , you need to subtract $\pi$ from the principal value.

When $xy = 1$ , either $x > 0, y > 0$ or $x < 0, y < 0$ . When $x > 0, y > 0$ , then $A = \tan^{-1}x, B = \tan^{-1}y \in (0, \pi/2)$ . The formula can be expressed as $\tan^{-1}x + \tan^{-1}y = \pi/2$ with $\tan^{-1}\frac{x+y}{1-xy}$ tending to infinity.

When $x < 0, y < 0$ , then $A = \tan^{-1}x, B = \tan^{-1}y \in (-\pi/2, 0)$ . The formula can be expressed as $\tan^{-1}x + \tan^{-1}y = -\pi/2$ with $\tan^{-1}\frac{x+y}{1-xy}$ tending to negative infinity.

$\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy}$

Derivation and Explanation:

Using the same substitutions, $\tan^{-1}x = A$ and $\tan^{-1}y = B$ , we have $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{x-y}{1+xy}$ . Therefore, $A-B = \tan^{-1}\frac{x-y}{1+xy}$ . No additional conditions are required here since the range of $\tan^{-1}z$ is $(-\pi/2, \pi/2)$ and $A-B$ will always fall within it, as long as $A$ and $B$ are in that range.

$2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}, |x| < 1$

Derivation and Explanation:

This is a special case of the addition formula. Let $y = x$ in the $\tan^{-1}x + \tan^{-1}y$ formula. We get $2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}$ . Similar to the condition $xy < 1$ , for this formula to be directly applicable, we need $x^2 < 1$ , which implies $|x| < 1$ . Always remember to check this condition before applying the formula. If $|x| > 1$ , you would need to consider additional cases involving adding or subtracting $\pi$ .

2. $\sin^{-1}x \pm \sin^{-1}y$ Formulas

$\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$

where $x,y \ge 0$ , $x^2 + y^2 \le 1$ or $x, y < 0$ , $x^2 + y^2 > 1$

$\sin^{-1}x - \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} - y\sqrt{1-x^2})$

where $x,y \ge 0$ , $x^2 + y^2 \le 1$ or $x, y < 0$ , $x^2 + y^2 > 1$

Derivation and Explanation:

Let $\sin^{-1}x = A$ and $\sin^{-1}y = B$ . Then, $x = \sin A$ and $y = \sin B$ . Using $\sin(A+B) = \sin A \cos B + \cos A \sin B = x\sqrt{1-y^2} + y\sqrt{1-x^2}$ . Therefore, $A+B = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$ . Be wary of the range restrictions on $x$ and $y$ above.

Similarly, $\sin(A-B) = \sin A \cos B - \cos A \sin B = x\sqrt{1-y^2} - y\sqrt{1-x^2}$ . Therefore, $A-B = \sin^{-1}(x\sqrt{1-y^2} - y\sqrt{1-x^2})$ .

3. $\cos^{-1}x \pm \cos^{-1}y$ Formulas

$\cos^{-1}x + \cos^{-1}y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$

where $x+y \ge 0$

$\cos^{-1}x - \cos^{-1}y = \cos^{-1}(xy + \sqrt{1-x^2}\sqrt{1-y^2})$

where $x \le y$

Derivation and Explanation:

Let $\cos^{-1}x = A$ and $\cos^{-1}y = B$ . Then, $x = \cos A$ and $y = \cos B$ . Using $\cos(A+B) = \cos A \cos B - \sin A \sin B = xy - \sqrt{1-x^2}\sqrt{1-y^2}$ . Therefore, $A+B = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ . Similarly, $\cos(A-B) = \cos A \cos B + \sin A \sin B = xy + \sqrt{1-x^2}\sqrt{1-y^2}$ . Therefore, $A-B = \cos^{-1}(xy + \sqrt{1-x^2}\sqrt{1-y^2})$ .

Tip: When in doubt, convert $\sin^{-1}x$ and $\cos^{-1}x$ to $\tan^{-1}$ expressions. The tangent formulas are often easier to work with. Use the right-angled triangle approach to find the corresponding tangent value.

Common Mistake: Forgetting to check the conditions for validity, especially for $\tan^{-1}x + \tan^{-1}y$ . This can lead to incorrect answers. Always verify if $xy < 1$ before directly applying the formula. Similarly, pay attention to the signs and quadrants.

JEE Trick: Look for patterns! Many JEE problems are designed to test your understanding of these formulas indirectly. Sometimes, the expression might not directly look like a standard formula, but with a bit of manipulation (like adding and subtracting terms or clever substitutions), you can transform it into a usable form.

By mastering these addition and subtraction formulas and being mindful of their conditions, you'll be well-equipped to tackle a wide range of inverse trigonometric function problems in the JEE Main exam. Keep practicing, and all the best!