Question

Please help me to prove this!​

Answer 1

Answer: see proof below

Explanation:

Use the following Identities in the proof.

Sum & Difference Identities:

`\cot (A+B)=(\cot A\cdot \cot B-1)/(\cot B+\cot A)\qquad \qquad \tan(A-B)=(\tan A-\tan B)/(1+\tan A\cdot \tan B)`

Half Angle Identities:

`\cot \bigg((A)/(2)\bigg)=(1+\cos A)/(\sin A)\qquad \qquad \qquad \tan\bigg((A)/(2)\bigg)=(1-\cos A)/(\sin A)`

Unit Circle:

`\cot \bigg((\pi)/(4)\bigg)=1\qquad \qquad \qquad \qquad \tan\bigg((\pi)/(4)\bigg)=1`

Proof LHS → RHS:

`\text{LHS:}\qquad \qquad \qquad \qquad \cot\bigg((\pi)/(4)+(\theta)/(2)\bigg)-\tan\bigg((\theta)/(2)-(\pi)/(4)\bigg)`

`\text{Sum and Difference:}\qquad (\cot ((\pi)/(4))\cdot \cot ((\theta)/(2))-1)/(\cot ((\theta)/(2))+\cot ((\pi)/(4)))-(\tan ((\theta)/(2))-\tan ((\pi)/(4)))/(1+\tan ((\theta)/(2))\cdot \tan((\pi)/(4)))`

`\text{Unit Circle:}\qquad \qquad (\cot ((\theta)/(2))-1)/(\cot ((\theta)/(2))+1)-(\tan ((\theta)/(2))-1)/(\tan ((\theta)/(2))+1)`

`\text{Half Angle:}\qquad \quad ((1+\cos \theta)/(\sin \theta)-1)/((1+\cos \theta)/(\sin \theta)+1)-((1-\cos \theta)/(\sin \theta)-1)/((1-\cos \theta)/(\sin \theta)+1)`

`=((1+\cos \theta)/(\sin \theta)-(\sin \theta)/(\sin \theta))/((1+\cos \theta)/(\sin \theta)+(\sin \theta)/(\sin \theta))-((1-\cos \theta)/(\sin \theta)-(\sin \theta)/(\sin \theta))/((1-\cos \theta)/(\sin \theta)+(\sin \theta)/(\sin \theta))`

`=(1+\cos \theta -\sin \theta)/((1+\cos \theta)+\sin \theta)-(1-\cos \theta -\sin \theta)/((1-\cos \theta)+\sin \theta)`

`\text{Simplify:}\\ (1+\cos \theta -\sin \theta)/((1+\cos \theta)+\sin \theta)\bigg(((1-\cos \theta)+\sin \theta)/(1-\cos \theta)+\sin \theta)\bigg)-(1-\cos \theta -\sin \theta)/((1-\cos \theta)+\sin \theta)\bigg((1+\cos \theta)+\sin \theta)/(1+\cos \theta)+\sin \theta)\bigg)`

`=(2\sin \theta \cdot \cos \theta)/(2\sin \theta(1+\sin \theta))-(-2\sin \theta \cdot \cos \theta)/(2\sin \theta (1+\sin \theta))`

`=(4\sin \theta \cdot \cos \theta)/(2\sin \theta(1+\sin \theta))`

`=(2\cos \theta)/(1+\sin \theta)`

LHS = RHS
`\checkmark`