Question

please help me


trigonometry question of grade 9 optional mathematics from Nepal ​

Answer 1

Answer:

it's a trigonometry lesson right

Answer 2

Given Question :-

Prove that

$\bf \: \dfrac{1 + {tan}^{2} \alpha \: {cosec}^{2} \beta }{1 + {tan}^{2} \alpha \: {cosec}^{2} \theta} = \dfrac{1 + {sin}^{2} \alpha \: {cosec}^{2} \beta }{1 + {sin}^{2} \alpha \: {cosec}^{2} \theta}$

$\large\underline{\bold{Solution :- }}$

Identities Used :-

$(1). \: \boxed{ \tt \: {sec}^{2}x - {tan}^{2}x = 1}$

$(2). \: \boxed{ \tt \: {cosec}^{2}x - {cot}^{2}x = 1}$

$(3). \: \boxed{ \tt \:cosx \: \times \: secx = 1 }$

CALCULATION :-

• Consider LHS

$\rm :\longmapsto\:\dfrac{1 + {tan}^{2} \alpha \: {cosec}^{2} \beta }{1 + {tan}^{2} \alpha \: {cosec}^{2} \theta}$

$\rm :\longmapsto\:\dfrac{ {sec}^{2} \alpha - {tan}^{2} \alpha + {tan}^{2} \alpha \: {cosec}^{2} \beta }{{sec}^{2} \alpha - {tan}^{2} \alpha + {tan}^{2} \alpha \: {cosec}^{2}\theta}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \tt \: \because \: {sec}^{2} \alpha - {tan}^{2} \alpha = 1}$

$\rm :\longmapsto\:\dfrac{{sec}^{2} \alpha + {tan}^{2} \alpha( {cosec}^{2} \beta - 1)}{{sec}^{2} \alpha + {tan}^{2} \alpha( {cosec}^{2}\theta - 1)}$

$\rm :\longmapsto\:\dfrac{{sec}^{2} \alpha + {tan}^{2} \alpha \: {cot}^{2} \beta}{{sec}^{2} \alpha + {tan}^{2} \alpha \: {cot}^{2}\theta}$

$\rm :\longmapsto\:\dfrac{\dfrac{1}{ {cos}^{2} \alpha } + \dfrac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } \: {cot}^{2} \beta }{\dfrac{1}{ {cos}^{2} \alpha } + \dfrac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } \: {cot}^{2}\theta}$

$\rm :\longmapsto\:\dfrac{1 + {sin}^{2} \alpha \: {cosec}^{2} \beta }{1 + {sin}^{2} \alpha \: {cosec}^{2}\theta}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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