frac{d}{dx} left[ left( frac{tan^2 2x - tan^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $y = \frac{	an^2 2x - 	an^2 x}{1 - 	an^2 2x 	an^2 x}$.Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have:$y = \frac{(	an 2x - 	an x)(

Vedclass · Accepted Answer

$\sec^2 x$

Vedclass · Answer

(C) Let $y = \frac{	an^2 2x - 	an^2 x}{1 - 	an^2 2x 	an^2 x}$.Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have:$y = \frac{(	an 2x - 	an x)(	an 2x + 	an x)}{(1 + 	an 2x 	an x)(1 - 	an 2x 	an x)}$.Using the tangent addition and subtraction formulas $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$ and $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$y = 	an(2x - x) 	an(2x + x) = 	an x 	an 3x$.Now,we need to find $\frac{d}{dx} [y \cdot \cot 3x]$.Substituting $y$,we get $\frac{d}{dx} [	an x 	an 3x \cdot \cot 3x]$.Since $	an 3x \cdot \cot 3x = 1$,the expression simplifies to $\frac{d}{dx} [	an x]$.Therefore,$\frac{d}{dx} [	an x] = \sec^2 x$.

$\frac{d}{dx} \left[ \left( \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} \right) \cot 3x \right] =$

Similar Questions

$\frac{d}{dt}(\tan t + t^2 \operatorname{cosech} t)$ is equal to

If $f(x) = mx + c$,$f(0) = 1$,and $f'(0) = 1$,then find the value of $f(2)$.

If $y = \sin^{-1} \left( x\sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^2} \right)$ and $\frac{dy}{dx} = \frac{1}{2\sqrt{x(1 - x)}} + p$,then $p =$

If $y = \sin^{98}(x) \cdot \cos^{39}(x)$,then find $\frac{dy}{dx}$.

The left-hand derivative of $f(x) = \{x\} \sin(\pi x)$ at $x = k$ ($k$ is an integer) is (where $\{ \}$ denotes the fractional part function).

Vedclass Test Series

Exam Paper Generator

Online Exam Module