Question

(CosecA-sinA)(secA-cosA)

Answer 1

Hi there !!

(CosecA-sinA)(secA-cosA)

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

=(1-sin²A/sinA)(1-cos²A/cosA)

=cos²A/sinA ×sin²A/cos {1-sin²A=cos²A and 1-cos²A=sin²A}

=cosA × sinA

Thankyou :)

Answer 2

Solution:

(CosecA-sinA)(secA-cosA)

☞( 1/Sin A - Sin A) ( 1/Cos A - Cos A )

☞( 1 - Sin²A/Sin A ) ( 1-Cos²A/Cos A )

☞( Cos²A/Sin A ) ( Sin²A/ Cos A)

☞ Cos²A/Sin A × Sin²A/ Cos A

☞ Cos A × Sin A

☞ Cos A . Sin A

Trigonometry Identities Used:

☞ Sin²A + Cos²A = 1

☞ Sin²A = 1 - Cos²A

☞ Cos²A = 1 - Sin²A