...sigh..another problem how shall I prove the following? $$ {\cot A\over1- \tan A} + {\tan A \over 1- \cot A} = 1 + \tan A + \cot A$$
so what now? the following's what I've done: $$\cot A - \cot^2 A + \tan A- \tan^2 A \over 2 - \tan A - \cot A$$
Asked
Active
Viewed 147 times
7
-
1try to avoid LCM for the denominator before simplification. Observe that $1-\cot A=-\frac{1-\tan A}{\tan A}$ – lab bhattacharjee May 08 '13 at 15:12
2 Answers
5
Let $\tan A=a\implies \cot A=\frac1a$
So, the problem reduces to
$$\frac{\frac1a}{1-a}+\frac a{1-\frac1a}=\frac1{a(1-a)}+\frac{a^2}{a-1}$$ $$=\frac1{a(1-a)}-\frac{a^2}{1-a}=\frac{1-a^3}{a(1-a)}=\frac{1+a+a^2}a=a+\frac1a+1$$
Alternatively,
$$\frac{\tan A}{1-\cot A}=\frac{\tan^2A}{\tan A-1} (\text{ multiplying the numerator & the denominator by }\tan A) $$
$$\implies \frac{\tan A}{1-\cot A}=-\frac{\tan^2A}{1-\tan A}$$
$$\text{So,} {\cot A\over1- \tan A} + {\tan A \over 1- \cot A}={\cot A\over1- \tan A} -\frac{\tan^2A}{1-\tan A}=\frac{\cot A-\tan^2A}{1-\tan A}=\frac{1-\tan^3A}{\tan A(1-\tan A)} (\text{ multiplying the numerator & the denominator by }\tan A) $$
$$=\frac{1+\tan A+\tan^2A}{\tan A}\text{ (assuming }1-\tan A\ne0)$$
$$=\cot A+1+\tan A$$
lab bhattacharjee
- 274,582
-
+1 for the first method that makes the problem clearer. (Also, don't you mean "assuming $\tan A ≠ 0$"?) – user123 May 08 '13 at 15:21
-
@Magtheridon96, no. Have you seen that I've cancelled $1-\tan A?$ – lab bhattacharjee May 08 '13 at 15:24
-
-
-
2
$$ \begin{align} \frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}&=\frac{\cos A\cot A - \sin A \tan A}{\cos A-\sin A} \\\\&=\frac{\frac{\cos^2 A}{\sin A} - \frac{\sin^2 A}{\cos A}}{\cos A-\sin A} \\\\&=\frac{\cos^3 A - \sin^3 A}{(\cos A \sin A)(\cos A-\sin A)} \\\\&=\frac{\cos^2 A + \sin^2 A + \sin A\cos A}{\cos A \sin A} \\\\&=1+\cot A+\tan A \end{align} $$
Milind Hegde
- 3,914