Question

❐ Brainly Moderators
❐ Brainly Stars
❐ Brainly Best Users

Prove that

Sin theta + Cos theta / Sin theta - Cos theta + Sin theta - Cos theta/Sin theta + Cos theta = 2Sec²theta/Tan²theta - 1​

Answer 1

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

Consider,

$\rm :\longmapsto\:\dfrac{sin\theta + cos\theta }{sin\theta - cos\theta } + \dfrac{sin\theta - cos\theta }{sin\theta + cos\theta }$

On taking LCM, we get

$\rm \:  =  \: \dfrac{ {(sin\theta + cos\theta )}^{2} + {(sin\theta - cos\theta )}^{2} }{(sin\theta + cos\theta )(sin\theta - cos\theta )}$

We know,

$\boxed{ \bf{ \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2} )}}$

and

$\boxed{ \bf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2}}}$

So, on using these Identities, we get

$\rm \:  =  \: \dfrac{2( {sin}^{2}\theta + {cos}^{2}\theta )}{ {sin}^{2}\theta - {cos}^{2}\theta}$

We know,

$\boxed{ \bf{ \: {sin}^{2} x + {cos}^{2} x = 1}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{2 \times 1}{ {sin}^{2} \theta - {cos}^{2} \theta }$

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

Now,

$\sf \: Divide \: numerator \: and \: denominator \: by \: {cos}^{2}x, \: we \: get$

$\rm \:  =  \: \dfrac{\dfrac{2}{ {cos}^{2} \theta } }{\dfrac{ {sin}^{2}\theta }{ {cos}^{2} \theta } - 1}$

We know,

$\boxed{ \bf{ \: \frac{1}{cos\theta } = sec\theta }}$

and

$\boxed{ \bf{ \: \frac{sin\theta }{cos\theta } = tan\theta }}$

So, using these, we get

$\rm \:  =  \: \dfrac{2 {sec}^{2} \theta }{ {tan}^{2} \theta - 1}$

Hence,

$\red{\rm :\longmapsto\:\dfrac{sin\theta + cos\theta }{sin\theta - cos\theta } + \dfrac{sin\theta - cos\theta }{sin\theta + cos\theta } = \dfrac{2 {sec}^{2}\theta }{ {tan}^{2} \theta - 1}}$

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