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

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

Question

(A) Given expression: $\frac{	an A}{1-\cot A} + \frac{\cot A}{1-	an A}$ Convert $	an A$ and $\cot A$ in terms of $\sin A$ and $\cos A$: $= \frac{\f

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

Similar Questions

Let $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ where $x \in R$ and $k \ge 1$. Then $f_4(x) - f_6(x)$ is equal to:

If $\sin A + \sin B = x$ and $\cos A + \cos B = y$,then $\sin(A + B) = $

The integer part of the number $\sum_{k=0}^{44} \frac{1}{\cos k^{\circ} \cos (k+1)^{\circ}}$ is

Let $\frac{\pi}{2} < \theta < \pi$ and $\cot \theta = -\frac{1}{2 \sqrt{2}}$. Then the value of $\sin (\frac{15 \theta}{2}) (\cos 8 \theta + \sin 8 \theta) + \cos (\frac{15 \theta}{2}) (\cos 8 \theta - \sin 8 \theta)$ is equal to:

One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module