Question

If a sin^2 x+b cos^2x=c show that tan^2x =c-b/a-c​

Answer 1

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

$\rm :\longmapsto\:a {sin}^{2}x \: + \: b {cos}^{2}x = c$

$\large\underline{\sf{To\:show - }}$

$\rm :\longmapsto\: {tan}^{2} x = \dfrac{c - b}{a - c}$

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

Given that,

$\rm :\longmapsto\:a {sin}^{2}x \: + \: b {cos}^{2}x = c$

$\boxed{ \bf{ \: Divide \: both \: sides \: by \: {cos}^{2}x}}$

$\rm :\longmapsto\:\dfrac{a {sin}^{2} x}{ {cos}^{2} x} + \dfrac{b {cos}^{2} x}{ {cos}^{2} x} = \dfrac{c}{ {cos}^{2} x}$

We know,

$\boxed{ \bf{ \: tanx = \frac{sinx}{cosx}}}$

and

$\boxed{ \bf{ \: secx = \frac{1}{cosx}}}$

$\rm :\longmapsto\:a {tan}^{2}x + b = c \: {sec}^{2}x$

$\rm :\longmapsto\:a {tan}^{2}x + b = c \:(1 + {tan}^{2}x)$

$\rm :\longmapsto\:a {tan}^{2}x + b = c \: + \: c{tan}^{2}x$

$\rm :\longmapsto\:a {tan}^{2}x - {ctan}^{2}x = c \: - \: b$

$\rm :\longmapsto\:{tan}^{2}x (a- c) = c \: - \: b$

$\rm :\longmapsto\: {tan}^{2}x = \dfrac{c - b}{a - c}$

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