Question

Hey... brainly users

Answer my question plz ___

Prove that
(SecA+CosecA)(SinA+CosA)= SecACosecA+2

Answer 1

LHS =( 1 / cos A + 1 / sin A) ( Sin A + cos A)
= ( sin A + cos A / sin A cos A) ( Sin A +
cosA)
= ( SinA + CosA) ^2 / ( sin A cos A)
= Sin^2 A + cos ^A + 2 Sin A cos A / SinA cos A.
= 1 + 2 SinA cos A. / Sin A Cos A
= 1 / SinA Cos. A + 2 Sin A. Cos A/ Sin A cos A
= Sec A CosecA + 2
= R. H. S
PROVED.

Answer 2

(SecA+CosecA)(SinA+CosA)
=SecASinA+SecACosA+CosecASinA+CosecACosA
=SinA/CosA+CosA*1/CosA+SinA*1/SinA+CosA/SinA
=SinA/CosA+1+1+CosA/SinA
=SinA/CosA+CosA/SinA+2
=sinA*SinA+CosA*CosA/SinACosA + 2
=1/SinACosA + 2
=CosecAsecA+2


hope it help you