Question

Prove that
$\frac{sin \: theta \: + cos \: theta}{sin \: theta \: - cos \: theta} + \frac{sin \: theta \: - cos \: theta}{sin \: theta \: + cos \: theta} = \frac{2 {sec}^{2} \: theta}{ {tan}^{2} - 1 }$
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Answer 1

Question :-

$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta +\cos\theta}=\frac{2\sec^{2}\theta }{ \tan^{2}\theta-1}$

Answer :-

(sinθ + cosθ)^2 + (sinθ - cosθ)^2/(sinθ)^2 - (cosθ)^2

sin^2θ+ cos^2 θ+ 2sinθcosθ+ sin^2 θ+ cos^2θ - 2sinθcosθ/ sin^2θ- cos^2θ

1 + 2sinθcosθ + 1 - 2sinθcosθ/ sin^2 θ- cos^2 θ

2/ sin^2 θ- cos^2θ

2/ cos^2 θ(sin^2θ/cos^2θ - 1)

2/cos^2θ(tan^2θ-1)

= 2sec^2θ/tan^2θ-1

Therefore, LHS = RHS

Hence proved !

Answer 2

Answer:

hi mates this is ur answer

Step-by-step explanation:

➡️➡️sinθ + cosθ)² + (sinθ - cosθ)²/(sinθ)² - (cosθ)²

➡️➡️sin²θ+ cos²θ+ 2sinθcosθ+ sin²θ+ cos²θ - 2sinθcosθ/ sin²θ- cos²θ

➡️➡️1 +2sinθcosθ + 1- 2sinθcosθ/sin² θ- cos² θ²/ sin²θ- cos²θ

➡️➡️2/ cos²θ(sin²θ/cos²θ - 1)

➡️➡️2/cos²θ(tan²θ-1)

✍✍= 2sec²θ/tan²θ-1

✍✍ LHS = RHS✍✍