sec^2(A) + cosec^2(A) = sec^2(A).cosec^2(A) @michele_liano

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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we have to apply these identities: \[\sec A = \frac{1}{{\cos A}},\quad \csc A = \frac{1}{{\sin A}}\]
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have u tried to solve the expression?? try converting the terms into sin nd cos form nd then take the LCM nd solve
left side becomes: \[\sec A + {\left( {\csc A} \right)^2} = \frac{1}{{\cos A}} + {\left( {\frac{1}{{\sin A}}} \right)^2} = ...\] please continue
i'm confused
I think that there is a typo into your original expression, please check
fixed
ok! so left side becomes: \[{\left( {\sec A} \right)^2} + {\left( {\csc A} \right)^2} = {\left( {\frac{1}{{\cos A}}} \right)^2} + {\left( {\frac{1}{{\sin A}}} \right)^2} = ...\]
ok m calling sinA = x nd cosA=y 1/y^2 +1/x^2 take LCM we get - (x^2 + y^2)/(xy)^2 we know that sin^2A+cos^2A = 1 therefore the numerator = 1 nd we r left with 1/(xy)^2 =sec^A cosec^A
sin^2(A)+cos^2(A)/cos(A)sin(A) = RHS
yes
1/sin(A)cos(A) = sec(A)cosec(A) = RHS
more precisely, it is: sin^2(A)+cos^2(A)/cos(A)^2 sin(A)^2 = RHS
ok!thanks...
i nedd more help *need

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