Question

2 sin theta cos theta minus cos theta upon 1 minus sin theta + sin squared theta minus cos squared theta is equal to COT theta prove​

Answer 1

Answer:

To Prove: $\frac{2\:sin\,\theta\:cos\,\theta-cos\,\theta}{1-sin\,\theta+sin^2\,\theta-cos^2\,\theta}=cot\,\theta$

we use the following result,

sin² x + cos² x = 1

Consider,

LHS

$=\frac{2\:sin\,\theta\:cos\,\theta-cos\,\theta}{1-sin\,\theta+sin^2\,\theta-cos^2\,\theta}$

$=\frac{cos\,\theta\:(2\:sin\,\theta-1)}{1-cos^2\,\theta+sin^2\,\theta-sin\,\theta}$

$=\frac{cos\,\theta\:(2\:sin\,\theta-1)}{sin^2\,\theta+sin^2\,\theta-sin\,\theta}$

$=\frac{cos\,\theta\:(2\:sin\,\theta-1)}{2\:sin^2\,\theta-sin\,\theta}$

$=\frac{cos\,\theta\:(2\:sin\,\theta-1)}{sin\,\theta\:(2\:sin\,\theta-1)}$

$=\frac{cos\,\theta}{sin\,\theta}$

$=cot\,\theta$

= RHS

Hence Proved.

Answer 2

Step-by-step explanation:

LHS = 2SinAcosA-CosA /1-sinA +sin²A-cos²A

=cosA (2sinA-1)/1-cos²A+sin²A-sinA

= cosA(2sinA-1)/sin²A+sin²A-sinA

= cosA(2sinA-1)/sinA(2sinA-1)

= cosA/sinA

=cotA

=RHS

Hence Proved.....