Question

prove that whole root under 1 - cos A by 1 + Cos A = sin A by 1 + Cos A​

Answer 1

Answer:

Step-by-step explanation:

Given :

To prove ,

$\sqrt{\frac{1-cosA}{1+cosA} } = \frac{sinA}{1+cosA}$

Proof :

We know that,

$sin^2A + cos^2A = 1$

⇒ $sin^2A = 1 - cos^2A$

LHS = $\sqrt{\frac{1-cosA}{1+cosA} }$

By multipying and dividing by $\sqrt{1+cosA}$

We get,

⇒ $\sqrt{\frac{1-cosA}{1+cosA} } \times \sqrt{\frac{1+cosA}{1+cosA}}$

⇒ $\frac{\sqrt{1-cosA}}{\sqrt{1+cosA} } \times \frac{\sqrt{1+cosA}}{ \sqrt{1+cosA}}$

⇒ $\frac{\sqrt{1-cosA} \times \sqrt{1+cosA} }{\sqrt{1+cosA} \times \sqrt{1+cosA} }$

⇒ $\frac{\sqrt{(1-cosA) \times (1+cosA)} }{\sqrt{(1+cosA) \times (1+cosA)} }$

⇒ $\frac{\sqrt{(1^2-cos^2A)} }{\sqrt{(1+cosA)^2} }$

⇒ $\frac{\sqrt{1-cos^2A} }{(1+cosA) }$

⇒ $\frac{\sqrt{sin^2A} }{(1+cosA) }$

⇒ $\frac{sinA }{(1+cosA) }$ = RHS

Hence, proved

Answer 2

Solution:

LHS = [(√1-cosA)/(√1+cosA)

Multiply numerator and denominator by

[(√1+cosA)/(√1+cosA) we get

= [√(1-cosA)(1+cosA)]/(√1+cosA)²

= √(1-cos²A)/(1+cosA)

= (√sinA)²/(1+cosA)

= sinA/(1+cosA)

=RHS

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