Question

solve it anybody who know​

Answer 1

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

Prove that,

$\rm :\longmapsto\:\dfrac{ {tan}^{2} \alpha }{ {cos}^{2} \beta } - \dfrac{ {tan}^{2} \beta }{ {cos}^{2} \alpha } = {tan}^{2} \alpha - {tan}^{2} \beta$

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

Consider LHS,

$\rm :\longmapsto\:\dfrac{ {tan}^{2} \alpha }{ {cos}^{2} \beta } - \dfrac{ {tan}^{2} \beta }{ {cos}^{2} \alpha }$

We know,

$\boxed{ \rm{ \frac{1}{cosx} = secx}}$

So, above expression can be rewritten as

$\rm \: = \: \: {tan}^{2} \alpha \: {sec}^{2} \beta - {tan}^{2} \beta \: {sec}^{2} \alpha$

We know,

$\boxed{ \rm{ {sec}^{2}x = 1 + {tan}^{2} x}}$

So, using this we get

$\rm \: = \: \: {tan}^{2} \alpha (1 + \: {tan}^{2} \beta ) - {tan}^{2} \beta (1 + \: {tan}^{2} \alpha )$

$\rm \: = \: \: {tan}^{2} \alpha \: + \: {tan}^{2} \alpha \: {tan}^{2} \beta - {tan}^{2} \beta - {tan}^{2} \alpha \: {tan}^{2} \alpha )$

$\rm \: = \: \: {tan}^{2} \alpha \: - \: {tan}^{2} \beta$

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