Please someone help me. It's about proving related to transformation of trigonometric ratios. Thank you!!

Answer:
see explanation
Step-by-step explanation:
Using sum to product identities
cos x - cos y = - 2sin([tex]\frac{x+y}{2}[/tex] )sin([tex]\frac{x-y}{2}[/tex] ) = 2sin([tex]\frac{x+y}{2}[/tex] )sin([tex]\frac{y-x}{2}[/tex] )
cos x + cos y = 2cos([tex]\frac{x+y}{2}[/tex] )cos([tex]\frac{x-y}{2}[/tex] )
Note that
sin10° = sin(90 - 10)° = cos80°
Thus
[tex]\frac{2sin(\frac{10+80}{2})sin(\frac{80-10}{2}) }{2cos(\frac{10+80}{2})cos(\frac{80-10}{2}) }[/tex] ← cancel 2 from numerator/ denominator
= [tex]\frac{sin45}{cos45}[/tex] × [tex]\frac{sin35}{cos35}[/tex]
= tan45° × tan35° { tan45° = 1 ]
= 1 × tan35°
= tan35° ← as required
Answer: see proof below
Step-by-step explanation:
* Use the following Co-function Identity: sin A = cos(90 - A)
* Use the following Sum to Product Identities:
cos x - cos y = 2 sin [(x + y)/2] · sin [(x - y)/2]
cos x + cos y = 2 cos [(x + y)/2] · cos [(x - y)/2]
* Use the Unit Circle to evaluate cos 45 = sin 45 = √2/2
Proof LHS → RHS
[tex]\text{LHS:}\qquad \qquad \qquad \qquad \quad \dfrac{\cos 10-\sin 10}{\cos 10+\sin 10}\\\\\\\text{Co-function Identity:}\qquad \dfrac{\cos 10 -\cos (90-10)}{\cos 10+\cos (90-10)}\\\\\\\text{Product to Sum Identity:}\quad \dfrac{2\sin (\frac{10+80}{2})\sin (\frac{80-10}{2})}{2\cos (\frac{10+80}{2})\cos (\frac{80-10}{2})}\\\\\\\text{Simplify:}\qquad \qquad \qquad \qquad \dfrac{2\sin (45)\sin (35)}{2 \cos (45)\cos (35)}[/tex]
[tex].\qquad \qquad \qquad \qquad \qquad = \dfrac{2(\frac{\sqrt2}{2}) \sin (35)}{2(\frac{\sqrt2}{2}) \cos(35)}\\\\\\.\qquad \qquad \qquad \qquad \qquad = \tan (35)[/tex]
LHS = RHS: tan 35 = tan 35 [tex]\checkmark[/tex]