Question

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

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Answer 1

Answer:

$\large\orange{\boxed{hey}}$

Step-by-step explanation:

Taking LHS:

= (secA+cosA) (secA-cosA)

= (sec²A -- cos²A)

= {(1+tan²A) - (1 - sin²A)}

= ( 1+tan²A - 1 + sin²A)

LHS = RHS

$\large\pink{\boxed{\boxed{ tan²A+sin²A}}}$

$\large\red{\boxed{\underline{Hence\:proved.}}}$

$\large\green{\underline{Mark\:it\:as\: brainliest\:answer..}}$

Answer 2

Step-by-step explanation:

hope this helps you

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