Step-by-step explanation:
[tex]\large\underline{\sf{Given \:Question - }}[/tex]
[tex] \sf \: tan\theta = \dfrac{m}{n}, \: prove \: that \: \dfrac{msin\theta - ncos\theta}{msin\theta + ncos\theta} = \dfrac{ {m}^{2} - {n}^{2} }{ {m}^{2} + {n}^{2} } [/tex]
[tex] \green{\large\underline{\sf{Solution-}}}[/tex]
Given that
[tex]\red{\rm :\longmapsto\:tan\theta = \dfrac{m}{n} }[/tex]
Now, Consider
[tex]\rm :\longmapsto\:\dfrac{msin\theta - ncos\theta}{msin\theta + ncos\theta} [/tex]
[tex]\rm \: = \: \dfrac{cos\theta\bigg[m\dfrac{sin\theta}{cos\theta} - n\bigg]}{cos\theta\bigg[m\dfrac{sin\theta}{cos\theta} + n\bigg]} [/tex]
[tex]\rm \: = \: \dfrac{mtan\theta - n}{mtan\theta + n} [/tex]
[tex]\rm \: = \: \dfrac{m \times \dfrac{m}{n} - n}{m \times \dfrac{m}{n} + n} [/tex]
[tex]\rm \: = \: \dfrac{\dfrac{ {m}^{2} }{n} - n}{\dfrac{ {m}^{2} }{n} + n} [/tex]
[tex]\rm \: = \: \dfrac{\dfrac{ {m}^{2} - {n}^{2} }{n}}{\dfrac{ {m}^{2} + {n}^{2} }{n}} [/tex]
[tex]\rm \: = \: \dfrac{ {m}^{2} - {n}^{2} }{ {m}^{2} + {n}^{2} } [/tex]
Hence,
[tex] \red{\sf \: tan\theta = \dfrac{m}{n}, \: \rm \implies\: \: \dfrac{msin\theta - ncos\theta}{msin\theta + ncos\theta} = \dfrac{ {m}^{2} - {n}^{2} }{ {m}^{2} + {n}^{2} }}[/tex]
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Additional Information:-
Relationship between sides and T ratios
sin θ = Opposite Side/Hypotenuse
cos θ = Adjacent Side/Hypotenuse
tan θ = Opposite Side/Adjacent Side
sec θ = Hypotenuse/Adjacent Side
cosec θ = Hypotenuse/Opposite Side
cot θ = Adjacent Side/Opposite Side
Reciprocal Identities
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
sin θ = 1/cosec θ
cos θ = 1/sec θ
tan θ = 1/cot θ
Co-function Identities
sin (90°−x) = cos x
cos (90°−x) = sin x
tan (90°−x) = cot x
cot (90°−x) = tan x
sec (90°−x) = cosec x
cosec (90°−x) = sec x
Fundamental Trigonometric Identities
sin²θ + cos²θ = 1
sec²θ - tan²θ = 1
cosec²θ - cot²θ = 1
