Question

(1-cosA)÷sinA=sinA÷(1+cosA). Prove it.

Answer 1

Step-by-step explanation:

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Answer 2

Question:-

$\rm \frac{1 - \cos(A) }{ \sin(A) } = \frac{ \sin(A) }{1 + \cos(A) }$

Solution:-

$\rm \frac{1 - \cos(A) }{ \sin(A) } = \frac{ \sin(A) }{1 + \cos(A) }$

Applying cross multiplication

$\rm(1 - \cos(A) )(1 + \cos(A) ) = \sin(A) \times \sin(A)$

Using this identity

$\rm \: (a - b)(a + b) = ( {a}^{2} - {b}^{2} )$

We get

$\rm \:( 1 - \cos {}^{2} (A) ) = \sin {}^{2} (A)$

Apply trigonometry identities

$\rm \: \to \: \sin {}^{2} (A) + \cos {}^{2} (A) = 1$

$\rm \: \to 1 - \cos {}^{2} (A) = \sin {}^{2} (A)$

$\rm \: \to \: 1 - \sin {}^{2} (A) = \cos {}^{2} (A)$

We get

$\rm \:( 1 - \cos {}^{2} (A) ) = \sin {}^{2} (A)$

$\rm \: \sin { }^{2} (A) = \sin {}^{2} ( A)$

Hence proved

Some trigonometry identity

$\rm \: \to \: \sin {}^{2} (A) + \cos {}^{2} (A) = 1$

$\rm \: \to \: 1 + \tan {}^{2} (A) = \sec {}^{2} (A)$

$\rm \: \to \: 1 + \cot {}^{2} (A) = \csc {}^{2} (A)$

$\rm \to \: \tan(A) = \frac{ \sin(A) }{ \cos(A) }$

$\rm \to \: \cot(A) = \frac{ \cos(A) }{ \sin(A) }$

$\rm \: \to \: \tan(A) \times \cot(A) = 1$