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SIZIF [17.4K]
3 years ago
9

Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)

Mathematics
1 answer:
Allisa [31]3 years ago
3 0

Answer:

Given that we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

From trigonometric identities, we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}

1 -  sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}

1 +  sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

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