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

If $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin \left( \theta + \frac{4\pi}{3} \right)$,then:

If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\tan (\alpha+2 \beta)$ is equal to

If $3 \sin \alpha = 5 \sin \beta$,then $\tan \left(\frac{\alpha + \beta}{2}\right) \div \tan \left(\frac{\alpha - \beta}{2}\right) = $

If $\cos x - \sin x = \sqrt{a} \sin x$,then $a \sin x + \cos x - \sin x = $

If $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$,for $x \in R$,then $f(2002)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module