If y = tan^-1((3cos x - 4sin x)/(4cos x + 3sin x)) + 2tan^-1(x/(1+√(1-x^2))), then dy/dx at x = √3/2 is equal to:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #256

y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})

, then

\frac{d y}{d x}

x = \frac{\sqrt{3}}{2}

is equal to:

Clue 1 of 3: Simplify Inverse Trigonometric Expressions

Look for standard forms like $\tan^{- 1} (\frac{A - B}{1 + A B})$ and substitutions like $x = \sin θ$ to simplify the given inverse trigonometric functions.

🧠 Full Solution Path

Step 1: Simplify the first term of y✦ Active

Let $y_{1} = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x})$ . Divide the numerator and denominator by $4 \cos x$ :

y_{1} = \tan^{- 1} (\frac{\frac{3}{4} - \tan x}{1 + \frac{3}{4} \tan x})

This is of the form

\tan^{- 1} A - \tan^{- 1} B

. Let

A = \frac{3}{4}

and

B = \tan x

y_{1} = \tan^{- 1} (\frac{3}{4}) - \tan^{- 1} (\tan x) = \tan^{- 1} (\frac{3}{4}) - x

Now, differentiate

y_{1}

with respect to

x

\frac{d y_{1}}{d x} = 0 - 1 = - 1

Step 2: Simplify the second term of y○ Expand

Let $y_{2} = 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$ . Substitute $x = \sin θ$ . Since $x = \frac{\sqrt{3}}{2} > 0$ , we can assume $θ \in (0, \frac{π}{2})$ , so $\sqrt{1 - x^{2}} = \sqrt{1 - \sin^{2} θ} = \cos θ$ .

y_{2} = 2 \tan^{- 1} (\frac{\sin θ}{1 + \cos θ})

Using half-angle identities

\sin θ = 2 \sin (\frac{θ}{2}) \cos (\frac{θ}{2})

and

1 + \cos θ = 2 \cos^{2} (\frac{θ}{2})

y_{2} = 2 \tan^{- 1} (\frac{2 \sin (\frac{θ}{2}) \cos (\frac{θ}{2})}{2 \cos^{2} (\frac{θ}{2})}) = 2 \tan^{- 1} (\tan (\frac{θ}{2}))

Since

θ \in (0, \frac{π}{2})

, then

\frac{θ}{2} \in (0, \frac{π}{4})

, so

\tan^{- 1} (\tan (\frac{θ}{2})) = \frac{θ}{2}

y_{2} = 2 (\frac{θ}{2}) = θ

Since

x = \sin θ

, we have

θ = \sin^{- 1} x

. So

y_{2} = \sin^{- 1} x

. Now, differentiate

y_{2}

with respect to

x

\frac{d y_{2}}{d x} = \frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}

💡 Teacher's Secret Hint

Remember the domain restrictions for inverse trigonometric functions when simplifying expressions like $\tan^{- 1} (\tan A)$ or $\sin^{- 1} (\sin A)$ .

Step 3: Calculate the total derivative and evaluate at x○ Expand

The total derivative $\frac{d y}{d x}$ is the sum of the derivatives of $y_{1}$ and $y_{2}$ :

\frac{d y}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x} = - 1 + \frac{1}{\sqrt{1 - x^{2}}}

Now, substitute

x = \frac{\sqrt{3}}{2}

into the expression for

\frac{d y}{d x}

\frac{d y}{d x} |_{x = \frac{\sqrt{3}}{2}} = - 1 + \frac{1}{\sqrt{1 - {(\frac{\sqrt{3}}{2})}^{2}}}

\frac{d y}{d x} |_{x = \frac{\sqrt{3}}{2}} = - 1 + \frac{1}{\sqrt{1 - \frac{3}{4}}} = - 1 + \frac{1}{\sqrt{\frac{1}{4}}}

\frac{d y}{d x} |_{x = \frac{\sqrt{3}}{2}} = - 1 + \frac{1}{\frac{1}{2}} = - 1 + 2 = 1

💡 Teacher's Secret Hint

Ensure all algebraic simplifications are correct, especially with fractions and square roots.

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Maths Question 16 – JEE-MAIN 2026

Clue 1 of 3: Simplify Inverse Trigonometric Expressions

🧠 Full Solution Path