Question

If mcotθ=n,then find the value of msinθ-ncosθ\mcosθ+nsinθ​

Answer 1

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

Given that,

$\red{\rm :\longmapsto\: mcot\theta = n}$

$\bf\implies \:\boxed{ \tt{ \: cot \theta = \frac{n}{m} \: }}$

Now, Consider

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

can be rewritten as

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

We know,

$\boxed{ \tt{ \: cotx = \frac{cosx}{sinx} \: }}$

So, using this identity, we get

$\rm \: = \:\dfrac{m - ncot \theta}{mcot \theta + n}$

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

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

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

Therefore,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{msin \theta - ncos \theta}{mcos \theta + nsin \theta} =\dfrac{ {m}^{2} - {n}^{2} }{2mn} \: }}$

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

Answer 2

Answer:

Given that,

\red{\rm :\longmapsto\: mcot\theta = n}:⟼mcotθ=n

\bf\implies \:\boxed{ \tt{ \: cot \theta = \frac{n}{m} \: }}⟹

cotθ=

m

n

Now, Consider

\rm :\longmapsto\:\dfrac{msin \theta - ncos \theta}{mcos \theta + nsin \theta}:⟼

mcosθ+nsinθ

msinθ−ncosθ

can be rewritten as

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

sinθ[

sinθ

mcosθ

+n]

sinθ[m−

sinθ

ncosθ

]

We know,

\boxed{ \tt{ \: cotx = \frac{cosx}{sinx} \: }}

cotx=

sinx

cosx

So, using this identity, we get

\rm \: = \:\dfrac{m - ncot \theta}{mcot \theta + n}=

mcotθ+n

m−ncotθ

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

m×

m

n

+n

m−n×

m

n

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

mn+mn

m

2

−n

2

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

2mn

m

2

−n

2

Therefore,

\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{msin \theta - ncos \theta}{mcos \theta + nsin \theta} =\dfrac{ {m}^{2} - {n}^{2} }{2mn} \: }}:⟼

mcosθ+nsinθ

msinθ−ncosθ

=

2mn

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