Question

If cos alpha/cos beta=m and cos alpha and cos beta =n , show that (m²+n²)cos² beta=n²​

Answer 1

Appropriate Question :-

$\rm :\longmapsto\:If \: m = \dfrac{cos \alpha }{cos \beta } \: and \: n = \dfrac{cos \alpha }{sin \beta }$

show that

$\rm :\longmapsto\:( {m}^{2} + {n}^{2} ) {cos}^{2} \beta = {n}^{2}$

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

Given that

$\rm :\longmapsto\:\: m = \dfrac{cos \alpha }{cos \beta } \: and \: n = \dfrac{cos \alpha }{sin \beta }$

Consider,

$\rm :\longmapsto\:( {m}^{2} + {n}^{2} ) {cos}^{2} \beta$

$\rm \:  =  \:  \: \bigg\{ {\bigg(\dfrac{cos \alpha }{cos \beta } \bigg) }^{2} + {\bigg(\dfrac{cos \alpha }{sin \beta } \bigg) }^{2} \bigg \} {cos}^{2} \beta$

$\rm \:  =  \:  \: \bigg(\dfrac{ {cos}^{2} \alpha }{ {cos}^{2} \beta } + \dfrac{ {cos}^{2} \alpha }{ {sin}^{2} \beta } \bigg) {cos}^{2} \beta$

$\rm \:  =  \:  \: {cos}^{2} \alpha \: \bigg(\dfrac{1}{ {cos}^{2} \beta } + \dfrac{1}{ {sin}^{2} \beta } \bigg) {cos}^{2} \beta$

$\rm \:  =  \:  \: {cos}^{2} \alpha \: \bigg(\dfrac{ {sin}^{2} \beta + {cos}^{2} \beta }{ {cos}^{2} \beta \: {sin}^{2} \beta } \bigg) {cos}^{2} \beta$

$\red{\bigg \{ \because \tt \: {sin}^{2}x + {cos}^{2}x = 1 \bigg \}}$

$\rm \:  =  \:  \: \dfrac{ {cos}^{2} \beta }{ {sin}^{2} \alpha }$

$\rm \:  =  \:  \: {n}^{2}$

Hence, Proved

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