$\tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right) + \sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$ is equal to Options: (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{2}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{6}$

### Analyzing the Setup The expression provided for evaluation is: $$T = \tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right) + \sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$$ To solve this, we will simplify each term independently before combining them. ### Simplifying the First Term Let us focus on the first term: $\tan^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)$. Observe the denominator $3 + \sqrt{3}$. By treating $3$ as $(\sqrt{3})^2$, we can rewrite the expression as: $$\frac{1+\sqrt{3}}{(\sqrt{3})^2 + \sqrt{3}} = \frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3} + 1)}$$ Since the numerator $1 + \sqrt{3}$ and the factor $(\sqrt{3} + 1)$ are identical, they cancel out. This leaves us with: $$\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$ ### Simplifying the Second Term Now, consider the second term: $\sec^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$. Focusing on the fraction inside the square root, we factor the numerator and denominator: $$\frac{8+4\sqrt{3}}{6+3\sqrt{3}} = \frac{4(2+\sqrt{3})}{3(2+\sqrt{3})}$$ The binomial $(2 + \sqrt{3})$ cancels out, leaving us with $\sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$. We then evaluate the inverse secant: $$\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$$ ### Final Synthesis We now combine the results from the two phases: $$T = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6}$$ The final simplified value of the expression is: **$$\frac{\pi}{3}$$**

Wait, why do we rewrite $3$ as $(\sqrt{3})^2$ in the first term?

That is a fantastic question! In competitive exams like JEE, we often look for ways to factorize expressions to simplify them. By writing $3$ as $(\sqrt{3})^2$, we reveal that $3 + \sqrt{3} = \sqrt{3}(\sqrt{3} + 1)$. This allows us to cancel the $(1 + \sqrt{3})$ term in the numerator, turning a complex fraction into a simple constant. It is all about spotting the hidden structure!

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Animated Solution for Mathematics - Inverse Trigonometric Functions: $tan^{- 1} (\frac{1 + 3}{3 + 3}) + sec^{- 1} (\frac{8 + 4 3}{6 + 3 3})$ is equal to

Q: Is there a specific identity for $\sec^{-1}$ that I should know?

Absolutely! The most useful identity here is $\sec^{-1}(x) = \cos^{-1}(\frac{1}{x})$. When you encounter inverse secant, cosecant, or cotangent, it is almost always easier to convert them into their primary counterparts (cosine, sine, or tangent) using their reciprocal identities. It makes the evaluation much more intuitive.

Select Answer:

Visualized Solution

Introduction to the Expression

Given expression: $tan^{- 1} (\frac{1 + 3}{3 + 3}) + sec^{- 1} (\frac{8 + 4 3}{6 + 3 3})$
Objective: Simplify each term individually using algebraic manipulation.

Focusing on the First Term

Term 1: $tan^{- 1} (\frac{1 + 3}{3 + 3})$
Analyze the denominator: $3 + 3$

Factoring the Denominator

Rewrite $3$ as $(3)^{2}$
Denominator: $3 + 3 = 3 (3 + 1)$

Simplifying the First Argument

Substitute back: $\frac{1 + 3}{3 ( 3 + 1 )}$
Cancel the common term $(1 + 3)$
Result: $\frac{1}{3}$

Evaluating $tan^{- 1} (\frac{1}{3})$

Term 1 becomes: $tan^{- 1} (\frac{1}{3})$
Since $tan (\frac{π}{6}) = \frac{1}{3}$ , we have $tan^{- 1} (\frac{1}{3}) = \frac{π}{6}$

Analyzing the Second Term

Term 2: $sec^{- 1} (\frac{8 + 4 3}{6 + 3 3})$
Focus on the fraction inside the square root: $\frac{8 + 4 3}{6 + 3 3}$

Factoring the Numerator

Factor the numerator: $8 + 43 = 4 (2 + 3)$

Factoring the Denominator

Factor the denominator: $6 + 33 = 3 (2 + 3)$

Simplifying the Fraction

Inside the root: $\frac{4 ( 2 + 3 )}{3 ( 2 + 3 )}$
Cancel $(2 + 3)$ to get: $\frac{4}{3}$

Taking the Square Root

Apply the square root: $\frac{4}{3} = \frac{2}{3}$
Term 2 becomes: $sec^{- 1} (\frac{2}{3})$

Evaluating $sec^{- 1} (\frac{2}{3})$

Recall: $sec^{- 1} (x) = cos^{- 1} (\frac{1}{x})$
$sec^{- 1} (\frac{2}{3}) = cos^{- 1} (\frac{3}{2})$
Since $cos (\frac{π}{6}) = \frac{3}{2}$ , we have $sec^{- 1} (\frac{2}{3}) = \frac{π}{6}$

Summing the Results

Total Sum: $\frac{π}{6} + \frac{π}{6}$
Calculation: $\frac{2 π}{6} = \frac{π}{3}$

Final Conclusion

Final Answer: $\frac{π}{3}$
Correct Option: (3)

00:00 / 00:00

The Sigma Insight: Properties of Inverse Trigonometric Functions

Solution Diagram

Analyzing the Setup

The expression provided for evaluation is:

T = tan^{- 1} (\frac{1 + 3}{3 + 3}) + sec^{- 1} \frac{8 + 4 3}{6 + 3 3}

To solve this, we will simplify each term independently before combining them.

Simplifying the First Term

Let us focus on the first term:

tan^{- 1} (\frac{1 + 3}{3 + 3})

Observe the denominator

3 + 3

. By treating

3

(3)^{2}

, we can rewrite the expression as:

\frac{1 + 3}{( 3 ) ^{2} + 3} = \frac{1 + 3}{3 ( 3 + 1 )}

Since the numerator

1 + 3

and the factor

(3 + 1)

are identical, they cancel out. This leaves us with:

tan^{- 1} (\frac{1}{3}) = \frac{π}{6}

Simplifying the Second Term

Now, consider the second term:

sec^{- 1} (\frac{8 + 4 3}{6 + 3 3})

Focusing on the fraction inside the square root, we factor the numerator and denominator:

\frac{8 + 4 3}{6 + 3 3} = \frac{4 ( 2 + 3 )}{3 ( 2 + 3 )}

The binomial

(2 + 3)

cancels out, leaving us with

\frac{4}{3} = \frac{2}{3}

. We then evaluate the inverse secant:

sec^{- 1} (\frac{2}{3}) = cos^{- 1} (\frac{3}{2}) = \frac{π}{6}

Final Synthesis

We now combine the results from the two phases:

T = \frac{π}{6} + \frac{π}{6} = \frac{2 π}{6}

The final simplified value of the expression is:

$\frac{π}{3}$

Common Traps & FAQs

Wait, why do we rewrite

3

(3)^{2}

in the first term?

That is a fantastic question! In competitive exams like JEE, we often look for ways to factorize expressions to simplify them. By writing

3

(3)^{2}

, we reveal that

3 + 3 = 3 (3 + 1)

. This allows us to cancel the

(1 + 3)

term in the numerator, turning a complex fraction into a simple constant. It is all about spotting the hidden structure!

Is there a specific identity for

sec^{- 1}

that I should know?

Absolutely! The most useful identity here is

sec^{- 1} (x) = cos^{- 1} (\frac{1}{x})

. When you encounter inverse secant, cosecant, or cotangent, it is almost always easier to convert them into their primary counterparts (cosine, sine, or tangent) using their reciprocal identities. It makes the evaluation much more intuitive.

Do I need to worry about the domain of these inverse functions?

While it is always good practice to check the domain, in this specific problem, the arguments are well within the valid range for

tan^{- 1}

and

sec^{- 1}

. The focus here is on algebraic simplification, but never forget to keep an eye on the domain in more complex problems!

Animated Solution for Mathematics - Inverse Trigonometric Functions: $tan^{- 1} (\frac{1 + 3}{3 + 3}) + sec^{- 1} (\frac{8 + 4 3}{6 + 3 3})$ is equal to

Select Answer:

Visualized Solution

Introduction to the Expression

Focusing on the First Term

Factoring the Denominator

Simplifying the First Argument

Evaluating $tan^{- 1} (\frac{1}{3})$

Analyzing the Second Term

Factoring the Numerator

Factoring the Denominator

Simplifying the Fraction

Taking the Square Root

Evaluating $sec^{- 1} (\frac{2}{3})$

Summing the Results

Final Conclusion

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Simplifying the First Term

Simplifying the Second Term

Final Synthesis

Similar Questions

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Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1(3+3​1+3​​)+sec−1(6+33​8+43​​​) is equal to

Select Answer:

Visualized Solution

Introduction to the Expression

Focusing on the First Term

Factoring the Denominator

Simplifying the First Argument

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1(3​1​)

Analyzing the Second Term

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Taking the Square Root

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sec−1(3​2​)

Summing the Results

Final Conclusion

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Simplifying the First Term

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Final Synthesis

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Animated Solution for Mathematics - Inverse Trigonometric Functions: $tan^{- 1} (\frac{1 + 3}{3 + 3}) + sec^{- 1} (\frac{8 + 4 3}{6 + 3 3})$ is equal to

Evaluating $tan^{- 1} (\frac{1}{3})$

Evaluating $sec^{- 1} (\frac{2}{3})$

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