Question

Please help me the brainliest.

Answer 1

Answer:

A² - B² = 0 (Proved).

Step-by-step explanation:

Given that  

$A = \frac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta + 1}$

⇒$A = \frac{\csc \theta + \cot \theta - (\csc^{2} \theta - \cot^{2} \theta)}{\cot \theta - \csc \theta + 1}$

{Since, we know the identity as $1 = \csc^{2} \theta - \cot^{2} \theta$}

⇒ $A = \frac{\csc \theta + \cot \theta - (\csc \theta + \cot \theta)\times (\csc \theta - \cot \theta)}{\cot \theta - \csc \theta + 1}$

⇒ $A = \frac{(\csc \theta + \cot \theta) (\cot \theta - \csc \theta + 1)}{\cot \theta - \csc \theta + 1}$

⇒ $A = \csc \theta + \cot \theta$

Again, given that $B = \csc \theta + \cot \theta$

So, A = B . ⇒ (A - B) = 0.

Hence, A² - B² = (A + B)(A - B) = 0 (Proved)