Question

1+sec theta /sec theta = sin^2 theta / 1-cos theta.

Answer 1

Here is the answer for your question

Answer 2

Answer:

Proved

Step-by-step explanation:

 Consider the Given question,

$\frac{1+\sec A}{\sec A}=\frac{\sin^2 A}{1-\cos A}$

First consider the left hand side,

$\frac{1+\sec A}{\sec A}$

Converting sec A  into CosA  as $\sec A =\frac{1}{\cos A}$

Then, $\frac{1+\sec A}{\sec A}=\frac{1+ \frac{1}{\cos A}}{\frac{1}{\cos A}}$

Solving , we get,

$\frac{\frac{1+\cos A}{\cos A}}{{\frac{1}{\cos A}}}$

As both have same denominator, so gets cancelled , we are left with

LHS = $1+\cos A$

Now, consider The right hand side,

$\frac{\sin^2 A}{1-\cos A}$

Using identity $\sin^2 A= 1-\cos^2 A$

We get, $\frac{1-\cos^2 A}{1-\cos A}$

Apply identity , $a^2-b^2=(a+b)(a-b)$

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

Solving we get,

RHS =  $1+\cos A$

Thus, LHS  = RHS

Hence proved