Question

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Answer 1

Answer:

Step-by-step explanation:

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

multiplying and dividing by $\sqrt{\frac{1+cosA}{1+cosA} }$ will make no difference as it is equal to 1

therefore

$\sqrt{\frac{1-cosA}{1+cosA}$ x $\sqrt{\frac{1+cosA}{1+cosA} }$

we know that (x+y)(x-y) = $x^{2} - y^{2}$

therefore in the numerator and denominator we have

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

We also now that $1^{2} - cos^{2}A = sin^{2} A$          (trigonometric identity)

therefore

$\sqrt{\frac{sin^{2} A}{(1+cosA)^{2} }} }$

taking root

${\frac{sinA}{1+cosA} }} }$ = RHS

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Answer 2

Solution →

Solution is in the attachment so please check the attachment.

Formulas used →

• (a +b)(a-b) = a²-b²
• 1- cos²a = sin² a

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