Question

prove that Root over 1 + sin theta by 1 minus sin theta equals to sec theta + tan thetap​

Answer 1

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Answer 2

Question:

$\small{ \rm{Prove \: that \: \sqrt{ \frac{1 + sin \theta}{1 - sin \theta} } = sec \theta + tan \theta}}$

Answer:

Taking LHS & multiplying & dividing with 1 + sinθ

$\small{ \rm{ = \sqrt{ \frac{(1 + sin \theta)(1 + sin \theta)}{(1 + sin \theta)(1 - sin \theta)} } }}$

Using

(a + b)(a - b) = a² - b²

$= \sqrt{ \frac{(1 - \sin \theta) ^{2} }{ {1}^{2} - { \sin}^{2} \theta } }$

Using

1 - sin²θ = cos²θ

√a/b = √a/√b

$\small {\rm { \frac{\sqrt{(1 + sin \theta) ^{2} } }{ \sqrt{ {cos }^{2} \theta } } = \frac{1 + sin \theta}{cos \theta} }}$

We know that

a+b/2 = a/2 + b/2

${\rm{ = \frac{1}{cos \theta} + \frac{sin \theta}{cos \theta} }}$

Using

1/cosθ = secθ

sinθ/cosθ = tanθ

$\small{ \rm{ = sec \theta + tan \theta}}$

LHS = RHS

Hence proved

CONCEPTS USED:

→ Trigonometric ratios

→ Trigonometric identities