Question

please..... explain ​

Answer 1

See in the attachment .

Answer 2

To prove

$\large\mathtt{\sqrt{\frac{1 + sin\theta}{1 - sin\theta}} + \sqrt{\frac{1 - sin\theta}{1 + sin\theta}} = 2 secθ}$

Proof

$\large\mathtt{\implies\: \sqrt{\frac{1 + sin\theta}{1 - sin\theta}} + \sqrt{\frac{1 - sin\theta}{1 + sin\theta}} = 2 sec\theta}$

$\large\mathtt{\implies\: \frac{\sqrt{1 + sin\theta}}{\sqrt{1 - sin\theta}} + \frac{\sqrt{1 - sin\theta}}{\sqrt{1 + sin\theta}} = 2 sec\theta}$

Taking LCM

$\large\mathtt{\implies\: \frac{(\sqrt{1 + sin\theta})^2 + (\sqrt{1 - sin\theta})^2} { (\sqrt{1 - sin\theta})(\sqrt{1 + sin\theta})}= 2 sec\theta}$

Using {(a-b)(a+b) = a² - b²} in denominator

$\large\mathtt{\implies\: \frac{1 + sin\theta + 1 - sin\theta}{\sqrt{(1)^2 - (sin\theta)^2}}}$

$\large\mathtt{\implies\: \frac{1 + \cancel{sin\theta} + 1 - \cancel{sin\theta}}{\sqrt{(1)^2 - (sin\theta)^2}}}$

$\large\mathtt{\implies\: \frac{2}{\sqrt{1 - sin^2\theta}}}$

$\large\mathtt{\implies\: \frac{2}{\sqrt{cos^2\theta}}}$

$\large\mathtt{\implies\: \frac{2}{cos\theta}}$

$\large\mathtt{\implies\: 2 × \frac{1}{cos\theta}}$

$\large\mathtt{\implies\: 2 sec\theta}$

LHS = RHS

Hence proved

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Trigonometric identities used

✓ sin²θ + cos²θ = 1

▶ 1 - sin²θ = cos²θ

✓ 1/ cosθ = secθ