Prove that: $\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}=1+\sec\theta\,\text{cosec}\,\theta$.
Step-by-Step Solution
Key Concept: Convert everything to $\sin\theta$ and $\cos\theta$, combine over a common denominator, and simplify using $\sin^2\theta+\cos^2\theta=1$.
Write $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$. The first term becomes $\dfrac{\sin\theta/\cos\theta}{1-\cos\theta/\sin\theta}=\dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}$. [1.5 Marks]
Similarly, the second term becomes $\dfrac{\cos\theta/\sin\theta}{1-\sin\theta/\cos\theta}=\dfrac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)}=-\dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}$. [1.5 Marks]
Adding: $\dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}-\dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}=\dfrac{1}{\sin\theta-\cos\theta}\left(\dfrac{\sin^2\theta}{\cos\theta}-\dfrac{\cos^2\theta}{\sin\theta}\right)$. [1.0 Mark]
$=\dfrac{1}{\sin\theta-\cos\theta}\times\dfrac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta}$. Using $\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta)$, this becomes $\dfrac{\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}$. [0.75 Mark]
$=\dfrac{1}{\sin\theta\cos\theta}+1=\text{cosec}\,\theta\sec\theta+1=1+\sec\theta\,\text{cosec}\,\theta$. Hence proved. [0.25 Mark]
Correct Answer: