Question

Prove It. I'm stuck at it...​

Answer 1

Answer:

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

Let's work on the LHS first,

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

Now, let's rationalise the LHS;

$\implies \frac{\sqrt{1 - \sin \theta} }{\sqrt{1 + \sin \theta} } \times \frac{\sqrt{1 - \sin \theta} }{\sqrt{1 - \sin \theta} } \\ \\ \implies \frac{( \sqrt{1 - \sin \theta)^{2}}}{1 - \sin^{2} \theta } \\ \\ \implies \frac{1 - \sin \theta}{ \cos \theta}$

[Because, 1 - sin²θ = cos θ]

$\implies \dfrac{1}{ \cos \theta} - \dfrac{ \sin \theta}{cos \theta} \: \: \: ....(i)$

Now, as $\frac{1}{\cos \theta}$ =

secθ. And, $\frac{\sin \theta}{\cos \theta}$= tan θ.

So, we can write equation (i) as ;

⇒ Sec θ - tan θ

∴ LHS = RHS

Answer 2

Steps are provided in detail.

Hope it helped u✌️☺️