JEE Main 2013 (Offline)Question 1
The expression $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$ can be written as :
Trigonometric Ratio & Identities
Solution
Concept. Put $t=\tan A$; the sum becomes $\frac{t^3-1}{t(t-1)}$. Formulas used. (1) $t^3-1=(t-1)(t^2+t+1)$(2) $\tan A+\cot A=\dfrac{1}{\sin A\cos A}=\sec A\operatorname{cosec}A$Step 1 - Rewrite. With $t=\tan A$, $\cot A=\frac1t$: $$\frac{t}{1-\frac1t}+\frac{1/t}{1-t}=\frac{t^2}{t-1}-\frac{1}{t(t-1)}=\frac{t^3-1}{t(t-1)}.$$ Step 2 - Cancel. By (1), this is $\dfrac{t^2+t+1}{t}=t+1+\dfrac1t$. Step 3 - Simplify. By (2), $1+\tan A+\cot A=1+\sec A\operatorname{cosec}A$. Answer: (B)