Question

It's a proving sum. ch- trigonometry and pls no spam​

Answer 1

EXPLANATION.

$\implies \sqrt{\dfrac{1 - cos(A)}{1 + cos(A)} } \ = \dfrac{sin(A)}{1 + cos(A)}$

As we know that,

Rationalize the L.H.S of the equation, we get.

$\implies \sqrt{\dfrac{1 - cos(A)}{1 + cos(A)} \times \dfrac{1 + cos(A)}{1 +cos(A)} }$

$\implies \sqrt{\dfrac{1 - cos^{2}(A) }{(1 + cos(A))^{2} } }$

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

$\implies \dfrac{sin(A)}{1 + cos(A)}$

Hence proved.

                                                                                                                     

MORE INFORMATION.

(1) = sin²θ + cos²θ = 1.

(2) = 1 + tan²θ = sec²θ.

(3) = 1 + cot²θ = cosec²θ.

Answer 2

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$\dashrightarrow \sqrt{\dfrac{1 - cos(A)}{1 + cos(A)} } \ = \dfrac{sin(A)}{1 + cos(A)}$

By taking L.H.S

$\dashrightarrow \sqrt{\dfrac{1 - cos(A)}{1 + cos(A)} }$

Now multiply it by

${\dfrac{1 + cos(A)}{1 +cos(A)}}}$

$\dashrightarrow \sqrt{\dfrac{1 - cos(A)}{1 + cos(A)} \times \dfrac{1 + cos(A)}{1 +cos(A)} }$

$\dashrightarrow \sqrt{\dfrac{1 - cos^{2}(A) }{(1 + cos(A))^{2} } }$

$\dashrightarrow \sqrt{\dfrac{sin^{2}(A) }{(1 + cos(A))^{2} } }$

$\dashrightarrow \dfrac{sin(A)}{1 + cos(A)}$

Hence proved that,

$\dashrightarrow \sqrt{\dfrac{1 - cos(A)}{1 + cos(A)} } \ = \dfrac{sin(A)}{1 + cos(A)}$

                                                             

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