Algebraically prove the identity: `(sin^2(theta))/(1-cos(theta))=(sec(theta)+1)/sec(theta)`

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Lets consider the LHS

`(sin^2 theta)/(1-cos theta)`

`=> (1 - cos^2 theta)/(1-cos theta) `

`(sin^2 theta + cos^2 theta = 1)`

`=> ((1-cos theta)(1+cos theta))/(1-cos theta)`

`=> 1 + cos theta`

`=> 1 + (1/sec theta)`

`=> (sec theta +1 ) / sec theta`

` ` **hence LHS = RHS**

or we can consider the RHS as well

`(sec theta +1 ) / sec theta`

`=> ((1/cos theta) +1)/(1/cos theta)`

`=> 1 + cos theta`

`= 1 + cos theta X (1 - cos theta)/(1-cos theta)`

`=> (1- cos^2 theta)/(1-cos theta)`

`=> (sin^2 theta)/(1-cos theta)`

`hence RHS = LHS `

You have to do the algebraic proof. To do this i will use the pythagorean identities.

What you to prove is,

`(sin^2(theta))/(1-cos(theta)) = (sec(theta)+1)/sec(theta)`

Let ABC is a right angles triangle in standard notation with length c is being the hypotenuse. and theta is being the angle `BhatAC` .

Therefore,

`sin(theta) = a/c`

`cos(theta) = b/c`

`sec(theta) = c/b`

I will start from LHS,

`LHS = (sin^2(theta))/(1-cos(theta))`

`= (a/c)^2/(1-b/c)`

`= a^2/(c(c-b))`

But we know in a right angled triangle, `a^2+b^2 = c^2`

and `a^2 = c^2-b^2`

Therefore,

`LHS = (c^2-b^2)/(c(c-b))`

`= ((c+b)(c-b))/(c(c-b))`

`= (c+b)/c`

Dividing both numerator and denominator by b,

`LHS = ((c+b)/b)/(c/b)`

`= (c/b+1)/(c/b)`

According above ratios,

`LHS = (sec(theta)+1)/sec(theta)`

Therefore, LHS = RHS and,

`(sin^2(theta))/(1-cos(theta)) = (sec(theta)+1)/sec(theta)`

**Proved algebraically**

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