\tan ^ - 1( 1 + \sqrt 3 \over 3 + \sqrt 3 ) + \sec ^ - 1( \sqrt 8 + 4\sqrt 3 \over 6 + 3\sqrt 3 ) is equal to :

Inverse Trigonometric Functions PYQ — JEE Main 2023 · 24 Jan · Shift 1 · Q28 with Solution

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Correct answer: B

Let $E=\tan^{-1}\left(\frac{1+\sqrt3}{3+\sqrt3}\right)+\sec^{-1}\left(\sqrt{\frac{8+4\sqrt3}{6+3\sqrt3}}\right).$

We simplify both terms separately.

Rationalize/simplify the fraction:

\frac{(1+\sqrt3)(3-\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}.$$ Denominator: $$9-3=6.$$ Numerator: $$ (1+\sqrt3)(3-\sqrt3)=3-\sqrt3+3\sqrt3-3=2\sqrt3.$$ So, $$\frac{1+\sqrt3}{3+\sqrt3}=\frac{2\sqrt3}{6}=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}.$$ Hence, $$\tan^{-1}\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6}.$$ --- 3. Second term: $$\sec^{-1}\left(\sqrt{\frac{8+4\sqrt3}{6+3\sqrt3}}\right).$$ Factor numerator and denominator: $$8+4\sqrt3=4(2+\sqrt3), \qquad 6+3\sqrt3=3(2+\sqrt3).$$ Therefore, $$\frac{8+4\sqrt3}{6+3\sqrt3}=\frac{4(2+\sqrt3)}{3(2+\sqrt3)}=\frac{4}{3}.$$ So the inside becomes $$\sqrt{\frac{4}{3}}=\frac{2}{\sqrt3}.$$ Thus we need $$\sec^{-1}\left(\frac{2}{\sqrt3}\right).$$ Let $$\theta=\sec^{-1}\left(\frac{2}{\sqrt3}\right).$$ Then $$\sec\theta=\frac{2}{\sqrt3} \implies \cos\theta=\frac{\sqrt3}{2}.$$ In the principal range of $\sec^{-1}$, this gives $$\theta=\frac{\pi}{6}.$$ So, $$\sec^{-1}\left(\frac{2}{\sqrt3}\right)=\frac{\pi}{6}.$$ --- 4. Add both terms: $$E=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}.$$ --- 5. Compare with options: - A: $\frac{\pi}{2}$ - B: $\frac{\pi}{3}$ - C: $\frac{\pi}{6}$ - D: $\frac{\pi}{4}$ So the correct option is $$\boxed{\frac{\pi}{3}}$$ which is **Option B**.

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View written solutionFree

Correct answer: B

Let $E=\tan^{-1}\left(\frac{1+\sqrt3}{3+\sqrt3}\right)+\sec^{-1}\left(\sqrt{\frac{8+4\sqrt3}{6+3\sqrt3}}\right).$

We simplify both terms separately.

Rationalize/simplify the fraction:

\frac{(1+\sqrt3)(3-\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}.$$ Denominator: $$9-3=6.$$ Numerator: $$ (1+\sqrt3)(3-\sqrt3)=3-\sqrt3+3\sqrt3-3=2\sqrt3.$$ So, $$\frac{1+\sqrt3}{3+\sqrt3}=\frac{2\sqrt3}{6}=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}.$$ Hence, $$\tan^{-1}\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6}.$$ --- 3. Second term: $$\sec^{-1}\left(\sqrt{\frac{8+4\sqrt3}{6+3\sqrt3}}\right).$$ Factor numerator and denominator: $$8+4\sqrt3=4(2+\sqrt3), \qquad 6+3\sqrt3=3(2+\sqrt3).$$ Therefore, $$\frac{8+4\sqrt3}{6+3\sqrt3}=\frac{4(2+\sqrt3)}{3(2+\sqrt3)}=\frac{4}{3}.$$ So the inside becomes $$\sqrt{\frac{4}{3}}=\frac{2}{\sqrt3}.$$ Thus we need $$\sec^{-1}\left(\frac{2}{\sqrt3}\right).$$ Let $$\theta=\sec^{-1}\left(\frac{2}{\sqrt3}\right).$$ Then $$\sec\theta=\frac{2}{\sqrt3} \implies \cos\theta=\frac{\sqrt3}{2}.$$ In the principal range of $\sec^{-1}$, this gives $$\theta=\frac{\pi}{6}.$$ So, $$\sec^{-1}\left(\frac{2}{\sqrt3}\right)=\frac{\pi}{6}.$$ --- 4. Add both terms: $$E=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}.$$ --- 5. Compare with options: - A: $\frac{\pi}{2}$ - B: $\frac{\pi}{3}$ - C: $\frac{\pi}{6}$ - D: $\frac{\pi}{4}$ So the correct option is $$\boxed{\frac{\pi}{3}}$$ which is **Option B**.