Question

Please help me to prove this!​

Answer 1

Answer:  see proof below

Step-by-step explanation:

Use the following Identities in the proof.

Sum & Difference Identities:

$\cot (A+B)=\dfrac{\cot A\cdot \cot B-1}{\cot B+\cot A}\qquad \qquad \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\cdot \tan B}$

Half Angle Identities:

$\cot \bigg(\dfrac{A}{2}\bigg)=\dfrac{1+\cos A}{\sin A}\qquad \qquad \qquad \tan\bigg(\dfrac{A}{2}\bigg)=\dfrac{1-\cos A}{\sin A}$

Unit Circle:

$\cot \bigg(\dfrac{\pi}{4}\bigg)=1\qquad \qquad \qquad \qquad \tan\bigg(\dfrac{\pi}{4}\bigg)=1$

Proof LHS → RHS:

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

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

$\text{Unit Circle:}\qquad \qquad \dfrac{\cot (\frac{\theta}{2})-1}{\cot (\frac{\theta}{2})+1}-\dfrac{\tan (\frac{\theta}{2})-1}{\tan (\frac{\theta}{2})+1}$

$\text{Half Angle:}\qquad \quad \dfrac{\frac{1+\cos \theta}{\sin \theta}-1}{\frac{1+\cos \theta}{\sin \theta}+1}-\dfrac{\frac{1-\cos \theta}{\sin \theta}-1}{\frac{1-\cos \theta}{\sin \theta}+1}$

                        $=\dfrac{\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{\sin \theta}}{\frac{1+\cos \theta}{\sin \theta}+\frac{\sin \theta}{\sin \theta}}-\dfrac{\frac{1-\cos \theta}{\sin \theta}-\frac{\sin \theta}{\sin \theta}}{\frac{1-\cos \theta}{\sin \theta}+\frac{\sin \theta}{\sin \theta}}$

                        $=\dfrac{1+\cos \theta -\sin \theta}{(1+\cos \theta)+\sin \theta}-\dfrac{1-\cos \theta -\sin \theta}{(1-\cos \theta)+\sin \theta}$

$\text{Simplify:}\\ \dfrac{1+\cos \theta -\sin \theta}{(1+\cos \theta)+\sin \theta}\bigg(\dfrac{(1-\cos \theta)+\sin \theta}{1-\cos \theta)+\sin \theta}\bigg)-\dfrac{1-\cos \theta -\sin \theta}{(1-\cos \theta)+\sin \theta}\bigg(\dfrac{1+\cos \theta)+\sin \theta}{1+\cos \theta)+\sin \theta}\bigg)$

$=\dfrac{2\sin \theta \cdot \cos \theta}{2\sin \theta(1+\sin \theta)}-\dfrac{-2\sin \theta \cdot \cos \theta}{2\sin \theta (1+\sin \theta)}$

$=\dfrac{4\sin \theta \cdot \cos \theta}{2\sin \theta(1+\sin \theta)}$

$=\dfrac{2\cos \theta}{1+\sin \theta}$

LHS = RHS $\checkmark$