Question

(sinA/1-cosA - 1-cosA/sinA) * (cosA/1-sinA - 1-sinA/cosA)=4​

Answer 1

Answer:

Use basic method to solve this question.

In the expression, ( sinA/ 1-cosA - 1-cosA/sinA) and (cosA/ 1-sinA - 1-sinA/cosA), take LCM in each expression to solve the question.

In ( sinA/ 1-cosA - 1-cosA/sinA), take LCM as sinA. 1-cosA

Now, the expression becomes [sin²A - (1-cosA)²] / sinA. (1-cosA)

Now, [ sin²A - (1+ cos²A + 2cosA )] / sinA. (1-cosA)

= [ sin²A - 1-cos²A + 2cosA] / sinA. (1-cosA)

= [ sin²A - 1 - 1 + sin²A +2 cosA] / sinA. ( 1-cosA)

= [ 2 sin²A - 2 + 2 cosA] / sinA. ( 1-cosA)

= [ 2- 2cos²A - 2 + 2cosA] / sinA.(1-cosA)

= [ 2cos A ( 1 - cos A) /  sinA. ( 1-cosA)

=2cos A/ sinA.............................................................equation 1

( You know than sin²A + cos²A = 1

Hence, sin²A = 1- cos²A )

Now, solve second expression ( cos A/ 1-sinA - 1-sinA/cosA )

The LCM is cos A . ( 1-sinA)

The expression becomes [cos²A - (1 -sinA)²] / cos A.  ( 1 - sin A)

[ cos²A - 1 - sin²A + 2 sinA ] /cosA. ( 1-sinA)

=  [ 1 -sin²A - 1 - sin²A + 2sinA] /cosA. ( 1-sinA)

= [ -2 sin²A + 2 sinA] / cosA. (1-sinA)

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

= 2sinA /cos A ..................................................................equation 2

Now equation 1 * equation 2 ( as given in the question)

2 cos A/ sinA * 2 sin A/cos A

= 4

Now, you can easily see that LHS  = RHS

Hence, the problem is solved.