Question

(sec A+ cos A)(sec A-cos A) = tan^2A + sin^2 A
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Answer 1

Step-by-step explanation:

LHS

(secA+cos)(secA-cosA)

we know that. (a^2-b^2)=(a+b)(a-b)

(sec^2A-cos^2A

.... sec^2A=1+tan^2A

(1+tan^2A-cos^2A)

(tan^2A+1-cos^2A)

(tan^2A+ sin^2A) .......RHS

hope it's helpful