Question

If tanθ = m/n, prove that: (m sinθ - n cosθ)/(m sinθ + n cosθ) = (m² - n²)/(m² + n²)​

Answer 1

Explanation:

`\large\underline{\sf{Given \:Question - }}`

`\sf \: tan\theta = (m)/(n), \: prove \: that \: (msin\theta - ncos\theta)/(msin\theta + ncos\theta) = \frac{ {m}^(2) - {n}^(2) }{ {m}^(2) + {n}^(2) }`

`\green{\large\underline{\sf{Solution-}}}`

Given that

`\red{\rm :\longmapsto\:tan\theta = (m)/(n) }`

Now, Consider

`\rm :\longmapsto\:(msin\theta - ncos\theta)/(msin\theta + ncos\theta)`

`\rm \:  =  \: (cos\theta\bigg[m(sin\theta)/(cos\theta) - n\bigg])/(cos\theta\bigg[m(sin\theta)/(cos\theta) + n\bigg])`

`\rm \:  =  \: (mtan\theta - n)/(mtan\theta + n)`

`\rm \:  =  \: (m * (m)/(n) - n)/(m * (m)/(n) + n)`

`\rm \:  =  \: \frac{\frac{ {m}^(2) }{n} - n}{\frac{ {m}^(2) }{n} + n}`

`\rm \:  =  \: \frac{\frac{ {m}^(2) - {n}^(2) }{n}}{\frac{ {m}^(2) + {n}^(2) }{n}}`

`\rm \:  =  \: \frac{ {m}^(2) - {n}^(2) }{ {m}^(2) + {n}^(2) }`

Hence,

`\red{\sf \: tan\theta = (m)/(n), \: \rm \implies\: \: (msin\theta - ncos\theta)/(msin\theta + ncos\theta) = \frac{ {m}^(2) - {n}^(2) }{ {m}^(2) + {n}^(2) }}`

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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