Question

if tanø°=a/b, show that (sin2Ø-cos2Ø/sin2Ø+cos2Ø)=(a2-b2/A2+b2)​

Answer 1

Appropriate Question :-

$\sf\:If \: tan\phi = \dfrac{a}{b}, \: show \: that \: \dfrac{ {sin}^{2}\phi - {cos}^{2}\phi }{ {sin}^{2}\phi + {cos}^{2}\phi} = \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

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

Given that

$\rm :\longmapsto\:tan\phi = \dfrac{a}{b}$

Now, Consider

$\rm :\longmapsto\:\dfrac{ {sin}^{2}\phi -{cos}^{2}\phi}{ {sin}^{2}\phi+{cos}^{2} \phi }$

$\red{ \boxed{ \sf{ \:Divide \: numerator \: and \: denominator \: by \: {cos}^{2}\phi }}}$

$\rm \:  =  \: \dfrac{\dfrac{{sin}^{2}\phi}{ {cos}^{2}\phi}-\dfrac{ {cos}^{2} \phi }{{cos}^{2} \phi } }{\dfrac{ {sin}^{2} \phi }{ {cos}^{2} \phi } + \dfrac{ {cos}^{2} \phi }{ {cos}^{2}\phi }}$

$\rm \:  =  \: \dfrac{ {tan}^{2}\phi - 1 }{ {tan}^{2} \phi + 1}$

$\rm \:  =  \: \dfrac{\dfrac{ {a}^{2} }{ {b}^{2} } - 1}{\dfrac{ {a}^{2} }{ {b}^{2} } + 1}$

$\rm \:  =  \: \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

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