Question

Simplify cotø(tanø+cotø)​

Answer 1

Answer:

$\large\boxed{\cot\theta(\tan\theta+\cot\theta)=1+\cot^2\theta=\dfrac{1}{\sin^2\theta}=\csc^2\theta}$

Step-by-step explanation:

$\text{Use}\\\\\text{distributive property:}\ a(b+c)=ab+ac\\\cot\alpha\tan\alpha=1.\\\\======================\\\\\cot\theta(\tan\theta+\cot\theta)=(\cot\theta)(\tan\theta)+(\cot\theta)(\cot\theta)\\\\=1+\cot^2\theta\\\\\text{If you want next transformation, then use:}\\\\\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}\\\\\sin^2\alpha+\cos^2\alpha=1\\\\=======================$

$=1+\left(\dfrac{\cos\theta}{\sin\theta}\right)^2=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta}{\sin^2\theta}+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}\\\\=\dfrac{1}{\sin^2\theta}\\\\\text{If you want next transformation, then use:}\\\\\csc\alpha=\dfrac{1}{\sin\alpha}\\\\=\left(\dfrac{1}{\sin\theta}\right)^2=(\csc\theta)^2=\csc^2\theta$