frac{cot A}{1-tan A} + frac{tan A}{1-cot A} = | vedclass

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

Question

(B) Given expression: $\frac{\cot A}{1-	an A} + \frac{	an A}{1-\cot A}$Substitute $\cot A = \frac{\cos A}{\sin A}$ and $	an A = \frac{\sin A}{\cos

Vedclass · Accepted Answer

$1 + \sec A \operatorname{cosec} A$

Vedclass · Answer

(B) Given expression: $\frac{\cot A}{1-	an A} + \frac{	an A}{1-\cot A}$Substitute $\cot A = \frac{\cos A}{\sin A}$ and $	an A = \frac{\sin A}{\cos A}$:$= \frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}} + \frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}$$= \frac{\cos^2 A}{\sin A(\cos A-\sin A)} + \frac{\sin^2 A}{\cos A(\sin A-\cos A)}$$= \frac{\cos^2 A}{\sin A(\cos A-\sin A)} - \frac{\sin^2 A}{\cos A(\cos A-\sin A)}$$= \frac{1}{(\cos A-\sin A)} \left[ \frac{\cos^3 A - \sin^3 A}{\sin A \cos A} ight]$Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:$= \frac{(\cos A-\sin A)(\cos^2 A + \sin^2 A + \sin A \cos A)}{(\cos A-\sin A) \sin A \cos A}$$= \frac{1 + \sin A \cos A}{\sin A \cos A}$$= \frac{1}{\sin A \cos A} + 1$$= \sec A \operatorname{cosec} A + 1$

$\frac{\cot A}{1-\tan A} + \frac{\tan A}{1-\cot A} = ?$

Similar Questions

$\sinh (x+y) \cosh (x-y)$ is equal to

The number of ordered pairs $(x, y)$ satisfying the equations $\sin x + \sin y = \sin(x + y)$ and $|x| + |y| = 1$ is

The value of $\sum_{r=1}^{18} \cos^2(5r)^\circ$,where $x^\circ$ denotes the $x$ degree,is equal to

$\cos^3 110^{\circ} + \cos^3 10^{\circ} + \cos^3 130^{\circ} = $

Evaluate: $\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \cos \frac{\pi}{14} \cos \frac{3 \pi}{14} \cos \frac{5 \pi}{14}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module