Question

1+sin ø/1-sin ø under root​

Answer 1

Appropriate Question:-

Prove that:

$\sqrt{ \dfrac{1 + \sin \theta}{1 - \sin \theta} } = \sec \theta - \tan \theta$

Solution:

Consider LHS

$\implies \sqrt{ \dfrac{1 + \sin \theta}{1 - \sin \theta} }$

Rationalise the denominator.

$\implies \sqrt{ \dfrac{1 + \sin \theta}{1 - \sin \theta} \times \dfrac{1 + \sin \theta}{1 + \sin \theta} }$

$\implies \sqrt{ \dfrac{(1 + \sin \theta)^{2} }{ {(1 - \sin \theta)(1 + \sin \theta)}} }$

$\implies \sqrt{ \dfrac{{(1 + \sin \theta)}^{2}}{ 1 - \sin^{2} \theta } }$

$\implies \sqrt{ \dfrac{ {(1 + \sin \theta)}^{2} }{ \cos^{2} \theta}}$

$\implies\dfrac{1 + \sin \theta}{ \cos \theta }$

$\implies\dfrac{1}{\cos \theta } + \dfrac{ \sin \theta}{ \cos \theta}$

$\implies\sec \theta + \tan \theta$

∴ LHS = RHS proved

Formula used:

• (A + B)(A - B) = A² - B²
• 1 - sin²x = cos²x
• 1/cos x = sec x
• (sin x)/(cos x) = tan x