Q.Find $\dfrac{dy}{dx}$ of the function expressed in parametric form: $\sin x = \dfrac{2t}{1 + t^2},\ \tan y = \dfrac{2t}{1 - t^2}$.

Concept: Parametric differentiation — here a standard identity collapses everything.

Recognise the double-angle forms. Writing $t=\tan\phi$:

So $x=y$, and therefore …

The idea

The expressions $\frac{2t}{1+t^2}$ and $\frac{2t}{1-t^2}$ are the tangent double-angle building blocks. Spotting them turns a messy quotient-rule problem into one line, and the same answer also drops out of the plain parametric method.

Method 1 — recognise the identity

Let $t=\tan\phi$. Then

So $\sin x=\sin 2\phi\Rightarrow x=2\phi=2\tan^{-1}t$, and $\tan y=\tan 2\phi\Rightarrow y=2\phi=2\tan^{-1}t$ (principal values). Hence $x=y$ and $\dfrac{dy}{dx}=1$.

Method 2 — differentiate through $t$

Differentiate each relation with respect to $t$:

$\cos x\,\frac{dx}{dt}=\frac{2(1-t^2)}{(1+t^2)^2},\qquad \sec^2 y\,\frac{dy}{dt}=\frac{2(1+t^2)}{(1-t^2)^2}.$ …

Method: Recognising Double-Angle Substitutions Hidden in Inverse Trig Parametric Forms

Some parametric pairs are deliberately disguised versions of the tangent double-angle formulas. Recognising the disguise turns a painful quotient-rule differentiation into a one-line answer.

Steps

Step 1: Compare the given expressions to the standard double-angle formulas

Memorise the pattern: if $t = \tan\phi$, then

Whenever you see $\dfrac{2t}{1+t^2}$ or $\dfrac{2t}{1-t^2}$ appearing as the argument of an inverse trig function (or as $\sin x$, $\tan y$, etc.), suspect this substitution immediately.

Step 2: Substitute $t = \tan\phi$ and simplify

Replace the given ratio with the matching double-angle expression, so each equation collapses to something like $\sin x = \sin 2\phi$ or $\tan y = \tan 2\phi$.

Step 3: Peel off the trig function to get $x$ and $y$ directly in terms of $\phi$ (hence $t$)

Using $\sin x = \sin 2\phi \Rightarrow x = 2\phi = 2\tan^{-1}t$ (within the principal range). Do this for both $x$ and $y$.

Step 4: Differentiate the now-simple expressions …

Common Mistakes

Mistake 1: Applying the quotient/implicit rule directly instead of spotting the double-angle pattern

Why it's wrong: differentiating $\sin x=\tfrac{2t}{1+t^2}$ and $\tan y=\tfrac{2t}{1-t^2}$ term-by-term with implicit differentiation is far messier than needed and invites algebra slips. Correct approach: recognize $\tfrac{2t}{1+t^2}=\sin2\phi$ and $\tfrac{2t}{1-t^2}=\tan2\phi$ for $t=\tan\phi$, collapsing both to $x=2\tan^{-1}t$ and $y=2\tan^{-1}t$.

Mistake 2: Skipping the principal-value check when writing $x=2\phi$, $y=2\phi$

Why it's wrong: $\sin x=\sin2\phi$ only forces $x=2\phi$ when $2\phi$ lies in the principal range of $\sin^{-1}$ (similarly for $\tan^{-1}$) — this is the single most-missed step on this exact NCERT Exemplar question. Correct approach: always state the range check even when it happens to work out, since the identity is not automatic. …