MHB Can you prove this identity using trigonometric identities?

[Drain Brain]

Help me get started with these problems.


Prove the following$\frac{\tan(A)}{1-\cot(A)}+\frac{\cot(A)}{1-\tan(A)}=\sec(A)\csc(A)+1$

$\frac{\sec(A)-\tan(A)}{\sec(A)+\tan(A)}=1-2\sec(A)\tan(A)+2\tan^{2}(A)$

 

[Siron]

Hint for 2: Multiply nominator and denominator with $(\sec(A)-\tan(A))$

 

[Drain Brain]

Hint for 2: Multiply nominator and denominator with $(\sec(A)-\tan(A))$

Actually, I'm done with prob 2. I need hint for prob 1. Thanks!

 

[soroban]

Hello, Drain Brain!

Prove: .$\frac{\tan A}{1-\cos A} + \frac{\cot A}{1-\tan A} \:=\:\sec A\csc A+1$

$\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A} \:=\: \dfrac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \dfrac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}} $

Multiply both fractions by $\frac{\sin A\cos A}{\sin A\cos A}\!:$

$\quad \frac{\sin^2\!A}{\sin A\cos A-\cos^2\!A} + \frac{\cos^2\!A}{\sin A \cos A - \sin^2\!A} $

$\quad =\;\frac{\sin^2\!A}{\cos A(\sin A - \cos A)} + \frac{\cos^2\!A}{-\sin A(\sin A - \cos A)} $

$\quad =\;\frac{\sin^3\!A}{\sin A\cos A(\sin A - \cos A)} - \frac{\cos^3\!A}{\sin A\cos A(\sin A - \cos A)}$

$\quad =\;\frac{\sin^3\!A\,-\,\cos^3\!A}{\sin A\cos A(\sin A\,-\,\cos A)} \;=\;\frac{(\sin A\,-\,\cos A)(\sin^2\!A\,+\,\sin A\cos A\,+\,\cos^2\!A)}{\sin A\cos A(\sin A\,-\,\cos A)}$

$\quad =\;\frac{(\sin^2\!A\,+\,\cos^2\!A)\,+\,\sin A\cos A}{\sin A\cos A} \;=\; \frac{1\,+\,\sin A\cos A}{\sin A\cos A}$

$\quad =\; \frac{1}{\sin A\cos A}\,+\,\frac{\sin A\cos A}{\sin A\cos A} \;=\; \sec A\csc A\,+\,1 $

 

[Greg]

$$\frac{\tan(A)}{1-\cot(A)}+\frac{\cot(A)}{1-\tan(A)}=\frac{\tan^2(A)-\cot(A)}{\tan(A)-1}\Leftrightarrow\frac{u}{1-\frac1u}+\frac{\frac1u}{1-u}$$$$=\frac{\cot(A)(\tan^3(A)-1)}{\tan(A)-1}$$$$=\frac{\cot(A)(\tan(A)-1)(\tan^2(A)+\tan(A)+1)}{\tan(A)-1}$$$$=\cot(A)(\tan^2(A)+\tan(A)+1)$$$$=\tan(A)+1+\cot(A)$$$$=\frac{1}{\cos(A)\sin(A)}+1$$$$=\sec(A)\csc(A)+1$$

 

[alyafey22]

Nothing more , but you might be interested ,

Let $\sin = u \,\,\,\,\, \cos = v$

$$\frac{\frac{u}{v}}{1-\frac{v}{u}}+\frac{\frac{v}{u}}{1-\frac{u}{v}} = \frac{u^2}{v(u-v)}-\frac{v^2}{u(u-v)}=\frac{u^3-v^3}{uv(u-v)} = \frac{1+uv}{uv} = \frac{1}{uv}+1$$