Question

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

Answer 1

Step-by-step explanation:

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

$\sf \: tan\theta = \dfrac{m}{n}, \: prove \: that \: \dfrac{msin\theta - ncos\theta}{msin\theta + ncos\theta} = \dfrac{ {m}^{2} - {n}^{2} }{ {m}^{2} + {n}^{2} }$

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

Given that

$\red{\rm :\longmapsto\:tan\theta = \dfrac{m}{n} }$

Now, Consider

$\rm :\longmapsto\:\dfrac{msin\theta - ncos\theta}{msin\theta + ncos\theta}$

$\rm \: = \: \dfrac{cos\theta\bigg[m\dfrac{sin\theta}{cos\theta} - n\bigg]}{cos\theta\bigg[m\dfrac{sin\theta}{cos\theta} + n\bigg]}$

$\rm \: = \: \dfrac{mtan\theta - n}{mtan\theta + n}$

$\rm \: = \: \dfrac{m \times \dfrac{m}{n} - n}{m \times \dfrac{m}{n} + n}$

$\rm \: = \: \dfrac{\dfrac{ {m}^{2} }{n} - n}{\dfrac{ {m}^{2} }{n} + n}$

$\rm \: = \: \dfrac{\dfrac{ {m}^{2} - {n}^{2} }{n}}{\dfrac{ {m}^{2} + {n}^{2} }{n}}$

$\rm \: = \: \dfrac{ {m}^{2} - {n}^{2} }{ {m}^{2} + {n}^{2} }$

Hence,

$\red{\sf \: tan\theta = \dfrac{m}{n}, \: \rm \implies\: \: \dfrac{msin\theta - ncos\theta}{msin\theta + ncos\theta} = \dfrac{ {m}^{2} - {n}^{2} }{ {m}^{2} + {n}^{2} }}$

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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