Question

Prove that
Sin theta + Cos theta / Sin theta - Cos theta + Sin theta - Cos theta/Sin theta + Cos theta = 2Sec²theta/Tan²theta - 1

Answer 1

TO PROVE :–

$\\ \bf \implies \dfrac{\sin \theta + \cos \theta}{ \sin \theta - \cos\theta} + \dfrac{\sin \theta - \cos\theta}{ \sin \theta + \cos\theta}= \dfrac{2 \sec ^{2} \theta}{ \tan^{2} \theta - 1}$

SOLUTION :–

• Let's take L.H.S. –

$\\ \bf \: \: = \: \: \dfrac{\sin \theta + \cos \theta}{ \sin \theta - \cos\theta} + \dfrac{\sin \theta - \cos\theta}{ \sin \theta + \cos\theta}$

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

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

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

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

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

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

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

$\\ \bf \: \: = \: \: \dfrac{2(1)}{\sin ^{2} \theta - \cos^{2} \theta}$

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

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

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

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

$\\ \bf \: \: = \: \: R.H.S.$

$\\ \bf \longrightarrow \: \: \green{\underbrace{ \red{Hence \: \: Proved}}}$

Answer 2

$\sf{Answer}$

Step by step explanation:-

We have to prove that LHS=RHS

So, LHS is not in simplest that means we can simplify and we can convert interms of RHS

So,take LHS and simplify

LHS= $\sf\dfrac{sinθ+cosθ}{sinθ-cosθ}$ +$\sf\dfrac{sinθ-cosθ}{sinθ+cosθ}$

Taking LCM for denominators

$\sf\dfrac{(sinθ+cosθ)²+(sinθ-cosθ)²}{(sinθ-cosθ)(sinθ+cosθ)}$

Numerator it is in form of

(a+b)² +(a-b)² = 2a²+2b²

Denominator in form of

(a+b)(a-b) = a²-b²

Similarily by using this identity we can simplify

$\sf\dfrac{2sin²θ+2cos²θ}{sin²θ-cos²θ}$

$\sf\dfrac{2(sin²θ+cos²θ)}{sin²θ-cos²θ}$

We know that

sin²θ+cos²θ = 1

$\sf\dfrac{2(1)}{sin²θ-cos²θ}$

Divide the denominator and numerator with cos²θ because in RHS tan²θ is there

$\sf\dfrac{2}{ sin²θ/cos²θ -cos²θ/cos²θ}$

We know that

sin²θ/cos²θ = tan²θ

$\sf\dfrac{2/cos²θ}{tan²θ -1}$

$\sf\dfrac{2sec²θ}{tan²θ -1}$

Hence proved

LHS = RHS

Hope my answer helps to u

Thank u :)