Question

Simplify cotø(tanø+cotø)​

Answer 1

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

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=(\cos\alpha)/(\sin\alpha)\\\\\sin^2\alpha+\cos^2\alpha=1\\\\=======================`

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