Write the differential coefficient of $\tan^{-1}\left(\dfrac{\sin x+\cos x}{\cos x-\sin x}\right)$ with respect to $x$.

> > [!TLDR] > The expression inside simplifies to $\tan\left(\frac{\pi}{4}+x\right)$, so the whole function reduces to $\frac{\pi}{4}+x$, whose derivative is $1$. > > Divide numerator and denominator of $\dfrac{\sin x+\cos x}{\cos x-\sin x}$ by $\cos x$: > $$\dfrac{\sin x+\cos x}{\cos x-\sin x}=\dfrac{1+\tan x}{1-\tan x}=\tan\left(\dfrac{\pi}{4}+x\right).$$ > …

Write the differential coefficient of tan⁻¹((sin x+cos x)/(cos x-sin x)) with respect to x.

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\tan^{-1}\left(\dfrac{\sin x+\cos x}{\cos x-\sin x}\right)

Odisha ChseOdisha CHSE +2 Science Board Exam 2020Subjective· 1mImportance★★★★★

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Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The inverse tangent function obeys a family of addition and doubling identities that let you combine two arctangents into one. They come straight from the tangent addition formula, but they carry conditions you must respect.

The core addition identity

Start from $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ . Put $A=\tan^{-1}x$ and $B=\tan^{-1}y$ , so $\tan A=x$ and $\tan B=y$ . Then

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\frac{x+y}{1-xy}\right), \qquad xy<1.

The restriction $xy<1$ keeps the combined angle inside the principal range $(-\pi/2,\pi/2)$ .

Watch out

If $xy>1$ the raw formula lands in the wrong branch, so you must correct it:

$\tan^{-1}x+\tan^{-1}y = \pi + \tan^{-1}\!\left(\frac{x+y}{1-xy}\right) \ (x,y>0),$

and $-\pi+\tan^{-1}(\cdot)$ when $x,y<0$ . Ignoring this is the classic exam slip.

Subtraction

Replacing $y$ with $-y$ gives

\tan^{-1}x - \tan^{-1}y = \tan^{-1}\!\left(\frac{x-y}{1+xy}\right), \qquad xy>-1.

The doubling identity

Set $y=x$ in the addition formula:

2\tan^{-1}x = \tan^{-1}\!\left(\frac{2x}{1-x^2}\right), \qquad -1<x<1.

The same angle can also be rewritten through sine and cosine, which is handy in integration and in proofs -- but each alternate form only matches $2\tan^{-1}x$ on part of its domain, so the two forms carry different conditions:

2\tan^{-1}x = \sin^{-1}\!\left(\frac{2x}{1+x^2}\right), \qquad -1\le x\le 1,

2\tan^{-1}x = \cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right), \qquad x\ge 0.

Watch out

The $\cos^{-1}$ form needs $x\ge 0$ -- it fails for negative $x$ . Check $x=-1$ : $2\tan^{-1}(-1)=2\left(-\tfrac{\pi}{4}\right)=-\tfrac{\pi}{2}$ , but $\cos^{-1}\!\left(\tfrac{1-1}{1+1}\right)=\cos^{-1}(0)=\tfrac{\pi}{2}$ , the wrong sign entirely. The $\sin^{-1}$ form has no such restriction because $\sin^{-1}$ (unlike $\cos^{-1}$ ) can return a negative angle.

The complementary identity

For every real $x$ ,

\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.

This holds without restriction because $\tan^{-1}$ and $\cot^{-1}$ of the same value are complementary angles. …

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Q.Write the differential coefficient of tan⁡−1(sin⁡x+cos⁡xcos⁡x−sin⁡x)\tan^{-1}\left(\dfrac{\sin x+\cos x}{\cos x-\sin x}\right)tan−1(cosx−sinxsinx+cosx​) with respect to xxx.

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

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Q.Write the differential coefficient of $\tan^{-1}\left(\dfrac{\sin x+\cos x}{\cos x-\sin x}\right)$ with respect to $x$ .

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