Considering the principal values of inverse trigonometric functions, the value of the expression tan

Q: Why do we convert everything to $\tan$? Can't we just use $\sin$ and $\cos$ identities?

You absolutely could, but think about the goal! The outer function is $\tan$. If you work with $\sin$ and $\cos$, you will end up with a mess of square roots and radicals. By converting to $\tan$ early, you keep the algebra clean and rational. It is about working smarter, not harder.

Q: What if the value inside the inverse trig function was negative? Would the triangle method still work?

Great question! The triangle method is a geometric tool. If the input is negative, you would need to account for the quadrant (e.g., $\sin^{-1}(-x) = -\sin^{-1}(x)$). Since our inputs here are positive, we stay in the first quadrant, making our lives much easier. Always check the sign before drawing your triangle!

Q: Is there any risk of the denominator becoming zero in the $\tan(A-B)$ formula?

In a general case, yes, you must check for domain restrictions. However, for these specific values, the denominator $1 + \tan A \tan B$ is clearly positive. Always keep an eye on the domain, but do not let the fear of undefined values paralyze your progress!

Question

Considering the principal values of inverse trigonometric functions, the value of the expression $	an(2 \sin^{-1}(\frac{2}{\sqrt{13}}) - 2 \cos^{-1}(\frac{3}{\sqrt{10}}))$ is equal to :

Options:

(A) $-\frac{16}{63}$

(B) $-\frac{33}{56}$

(C) $\frac{16}{63}$

(D) $\frac{33}{56}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a trigonometric expression; we are embarking on a journey to simplify complexity. Look at the problem: $	an(2 \sin^{-1}(\frac{2}{\sqrt{13}}) - 2 \cos^{-1}(\frac{3}{\sqrt{10}}))$.

It looks intimidating, but every complex expression is just a collection of simple parts waiting to be understood. Let us start by demystifying the inverse trigonometric functions.

We define $\alpha = \sin^{-1}(\frac{2}{\sqrt{13}})$. If $\sin \alpha = \frac{2}{\sqrt{13}}$, imagine a right-angled triangle where the opposite side is $2$ and the hypotenuse is $\sqrt{13}$.

By the Pythagorean theorem, the adjacent side is $\sqrt{(\sqrt{13})^2 - 2^2} = \sqrt{13 - 4} = 3$. Suddenly, the mystery vanishes, and we see that $	an \alpha = \frac{2}{3}$.

We do the exact same thing for $\gamma = \cos^{-1}(\frac{3}{\sqrt{10}})$. Here, the adjacent side is $3$ and the hypotenuse is $\sqrt{10}$. The opposite side becomes $\sqrt{(\sqrt{10})^2 - 3^2} = \sqrt{10 - 9} = 1$. Thus, $	an \gamma = \frac{1}{3}$.

### The Power of Double Angles

Now, look at the expression again. We have $2\alpha$ and $2\gamma$. We need to find the tangent of these doubled angles using the identity:

$$	an(2	heta) = \frac{2	an 	heta}{1 - 	an^2 	heta}$$

For our first angle, $\alpha$, we substitute $	an \alpha = \frac{2}{3}$:

$$	an(2\alpha) = \frac{2(\frac{2}{3})}{1 - (\frac{2}{3})^2} = \frac{\frac{4}{3}}{1 - \frac{4}{9}} = \frac{\frac{4}{3}}{\frac{5}{9}} = \frac{12}{5}$$

Now, for $\gamma$, we substitute $	an \gamma = \frac{1}{3}$:

$$	an(2\gamma) = \frac{2(\frac{1}{3})}{1 - (\frac{1}{3})^2} = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{3}{4}$$

We have transformed the original expression into a simple subtraction of two known values: $	an(2\alpha - 2\gamma)$.

### The Grand Finale: Compound Angles

We are at the final step. We need to evaluate $	an(A - B)$ where $A = 2\alpha$ and $B = 2\gamma$. The formula is:

$$	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$$

Substituting our values, we get:

$$\frac{\frac{12}{5} - \frac{3}{4}}{1 + (\frac{12}{5})(\frac{3}{4})}$$

Let us calculate the numerator: $\frac{48 - 15}{20} = \frac{33}{20}$. Now the denominator: $1 + \frac{36}{20} = \frac{20 + 36}{20} = \frac{56}{20}$.

When we divide these, the $20$s cancel out beautifully. The final result is **$\frac{33}{56}$**.

Animated Solution for Mathematics - Inverse Trigonometric Functions: Considering the principal values of inverse trigonometric functions, the value of the expression $tan (2 sin^{- 1} (\frac{2}{13}) - 2 cos^{- 1} (\frac{3}{10}))$ is equal to :

Select Answer:

Visualized Solution

Define $α = sin^{- 1} (\frac{2}{13})$

Analyze Triangle for $α$

Calculate $tan (2 α)$

Define $γ = cos^{- 1} (\frac{3}{10})$

Analyze Triangle for $γ$

Calculate $tan (2 γ)$

Apply $tan (A - B)$ Formula

Final Substitution and Calculation

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Power of Double Angles

The Grand Finale

Similar Questions

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Select Answer:

Visualized Solution

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=sin−1(13​2​)

Analyze Triangle for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(2α)

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ=cos−1(10​3​)

Analyze Triangle for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(2γ)

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(A−B) Formula

Final Substitution and Calculation

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Power of Double Angles

The Grand Finale

Similar Questions

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