Question

1 - cos theta by sin theta is equals to sin theta by 1 + cos theta prove it
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Answer 1

by taking RHS

rationalize 1 + cos theta

sin theta divided by 1 + cos theta x 1 - cos theta divided by 1 - cos thetatheta

sin theta x 1 - cos theta divided by one square minus cos square theta

sin theta x 1 - cos theta upon sin square theta

sin theta x 1 - cos theta upon sin theta into sin theta

1 - cos theta upon sin theta equal to RHS

Answer 2

Given :

$\dfrac{1 - \cos(\theta) }{ \sin(\theta) } = \dfrac{\sin(\theta)}{1 +\cos(\theta) }$

To prove :

LHS = RHS

Proof :

$\rm\implies LHS = \dfrac{1 - \cos(\theta) }{ \sin(\theta) }$

$\rm\implies \dfrac{1 - \cos(\theta) }{ \sin(\theta) } \times \dfrac{1 + \cos(\theta)}{1 + \cos(\theta)}$

we know that :- (a+b) (a-b) = a²-b²

$\rm\implies\dfrac{ {1}^{2} - {\cos}^{2} (\theta) }{ \sin(\theta)(1 + \cos\theta) }$

Now , 1 - cos²θ = sin² θ

$\rm\implies \dfrac{ {\sin}^{2}\theta }{ (\sin\theta)(1 + \cos\theta) }$

Now we will cancel out sin θ from numerator and denominator :-

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

Therefore, LHS = RHS.

Hence proved

Learn more:

sin²A + cos²A = 1

1 - sin²A = cos²A

1 - cos²A = sin²A

1 + tan²A = sec²A

sec²A - tan²A = 1

sec²A - 1 = tan²A

1 + cot²A = cosec²A

cosec²A - 1 = cot²A

cosec²A - cotA = 1

cotA = 1/tanA

cotA = cosA/sinA

Sin 30° = 1/2

cos 30° = √3/2

tan 30° = 1/√3

Sin 45° = 1/√2

cos 45° = 1/√2

tan 45° = 1

Sin 60° = √3/2

cos 60° = 1/2

tan 60° = √3

Sin 90° = 1

cos 90° = 0

tan 90° = infinite

cosec x = 1/ sin x

sec x = 1/ cosx

cot x = 1/tan x