Question

(SecA+cosA)(secA-cosA)=sin(^(2)A+tan(^(2)A

Answer 1

Solution:

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Here , we are using the following identities:

i ) (x+y)(x-y) = x²-y²

ii ) sec²A = 1+tan²A

iii) cos²A = 1-sin²A

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LHS =(secA+cosA)(secA-cosA)

= sec²A - cos²A

= (1+tan²A)-(1-sin²A)

= 1+ tan²A - 1 + sin²A

= sin²A + tan²A

= RHS

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Answer 2

Hey !

Refer the below attachment !

Thanks !