Answer:
Both sides = [tex]\frac{(sin\alpha+cos\alpha)}{(cos\alpha-sin\alpha)}[/tex]
Step-by-step explanation:
Let us revise some identity
sin(∝ + a) = sin(∝) cos(a) + cos(∝) sin(a)
cos(∝ + a) = cos(∝) cos(a) - sin(∝) sin(a)
tan ∝ = [tex]\frac{sin(\alpha)}{cos(\alpha)}[/tex]
Let us simplify each side
Left hand side
∵ [tex]tan(\alpha+\frac{\pi }{4})=\frac{sin(\alpha+\frac{\pi }{4})}{cos(\alpha+\frac{\pi}{4})}[/tex] ⇒ (1)
∵ [tex]sin(\alpha+\frac{\pi}{4})=sin(\alpha )cos(\frac{\pi}{4})+cos(\alpha)sin(\frac{\pi}{4})[/tex] ⇒ (2)
∵ [tex]sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}[/tex] and [tex]cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}[/tex]
- Substitute them in (2)
∴ [tex]sin(\alpha+\frac{\pi}{4})=sin(\alpha )(\frac{\sqrt{2}}{2})+cos(\alpha)(\frac{\sqrt{2} }{2})[/tex]
∴ [tex]sin(\alpha+\frac{\pi}{4})=(\frac{\sqrt{2}}{2})sin\alpha +(\frac{\sqrt{2} }{2})cos\alpha[/tex]
- Take [tex]\frac{\sqrt{2}}{2}[/tex] as a common factor
∴ [tex]sin(\alpha+\frac{\pi}{4})=(\frac{\sqrt{2}}{2})(sin\alpha +cos\alpha)[/tex] ⇒ (3)
∵ [tex]cos(\alpha+\frac{\pi}{4})=cos(\alpha )cos(\frac{\pi}{4})-sin(\alpha)sin(\frac{\pi}{4})[/tex] ⇒ (4)
∵ [tex]cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}[/tex] and [tex]sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}[/tex]
- Substitute them in (4)
∴ [tex]cos(\alpha+\frac{\pi}{4})=cos(\alpha )(\frac{\sqrt{2}}{2})-sin(\alpha)(\frac{\sqrt{2}}{2})[/tex]
∴ [tex]cos(\alpha+\frac{\pi}{4})=(\frac{\sqrt{2}}{2})cos\alpha-(\frac{\sqrt{2} }{2})sin\alpha[/tex]
- Take [tex]\frac{\sqrt{2}}{2}[/tex] as a common factor
∴ [tex]cos(\alpha+\frac{\pi}{4})=(\frac{\sqrt{2}}{2})(cos\alpha-sin\alpha)[/tex] ⇒ (5)
Substitute (3) and (5) in (1)
∴ [tex]tan(\alpha+\frac{\pi }{4})=\frac{\frac{\sqrt{2}}{2}(sin\alpha+cos\alpha)}{\frac{\sqrt{2}}{2}(cos\alpha-sin\alpha)}[/tex]
- Simplify it by divide up and down by [tex]\frac{\sqrt{2}}{2}[/tex]
∴ [tex]tan(\alpha+\frac{\pi }{4})=\frac{(sin\alpha+cos\alpha)}{(cos\alpha-sin\alpha)}[/tex]
∴ Left hand side = [tex]\frac{(sin\alpha+cos\alpha)}{(cos\alpha-sin\alpha)}[/tex]
Right hand side
Right hand side = [tex]\frac{cos(\alpha-\frac{5}{4}\pi )}{cos(\frac{3}{4}\pi-\alpha)}[/tex] ⇒ (1)
∵ [tex]cos(\alpha-\frac{5}{4}\pi)=cos(\alpha )cos(\frac{5}{4}\pi )+sin(\alpha)sin(\frac{5}{4}\pi)[/tex] ⇒ (2)
∵ [tex]cos(\frac{5}{4}\pi )=-\frac{\sqrt{2}}{2}[/tex] and [tex]sin(\frac{5}{4}\pi)=-\frac{\sqrt{2}}{2}[/tex]
- Substitute them in (2)
∴ [tex]cos(\alpha-\frac{5}{4}\pi)=cos(\alpha )(-\frac{\sqrt{2}}{2})+sin(\alpha)(-\frac{\sqrt{2}}{2})[/tex]
∴ [tex]cos(\alpha-\frac{5}{4}\pi )=(-\frac{\sqrt{2}}{2})cos\alpha+(-\frac{\sqrt{2} }{2})sin\alpha[/tex]
- Take [tex]-\frac{\sqrt{2}}{2}[/tex] as a common factor
∴ [tex]cos(\alpha-\frac{5}{4}\pi)=(-\frac{\sqrt{2}}{2})(cos\alpha+sin\alpha)[/tex] ⇒ (3)
∵ [tex]cos(\frac{3}{4}\pi-\alpha)=cos(\frac{3}{4}\pi)cos(\alpha)+sin(\frac{3}{4}\pi)sin(\alpha)[/tex] ⇒ (4)
∵ [tex]cos(\frac{3}{4}\pi )=-\frac{\sqrt{2}}{2}[/tex] and [tex]sin(\frac{3}{4}\pi)=\frac{\sqrt{2}}{2}[/tex]
- Substitute them in (4)
∴ [tex]cos(\frac{3}{4}\pi-\alpha)=(-\frac{\sqrt{2}}{2})cos(\alpha)+(\frac{\sqrt{2}}{2})sin(\alpha)[/tex]
- Take [tex]-\frac{\sqrt{2}}{2}[/tex] as a common factor
∴ [tex]cos(\frac{3}{4}\pi-\alpha)=(-\frac{\sqrt{2}}{2})(cos\alpha-sin\alpha)[/tex] ⇒ (5)
Substitute (3) and (5) in (1)
∴ [tex]\frac{cos(\alpha-\frac{5}{4}\pi )}{cos(\frac{3}{4}\pi-\alpha)}[/tex] = [tex]\frac{-\frac{\sqrt{2}}{2}(cos\alpha+sin\alpha)}{-\frac{\sqrt{2}}{2}(cos\alpha-sin\alpha)}[/tex]
- Simplify it by divide up and down by [tex]-\frac{\sqrt{2}}{2}[/tex]
∴ [tex]\frac{cos(\alpha-\frac{5}{4}\pi )}{cos(\frac{3}{4}\pi-\alpha)}[/tex] = [tex]\frac{(cos\alpha+sin\alpha)}{(cos\alpha-sin\alpha)}[/tex]
∴ Right hand side = [tex]\frac{(cos\alpha+sin\alpha)}{(cos\alpha-sin\alpha)}[/tex]
∵ cos∝ + sin∝ = sin∝ + cos∝
∴ Left hand side = Right hand side
∴ The identity is verified
