Question

Prove that :- sec2A+cosec2A=sec2A.cosec2A

Answer 1

HOlla mate,

Here ur answer.

LHS = sec²A +cosec²A

=> 1/cos²A + 1/sin²A

=> (sin²A +cos²A)/(cos²Asin²A)

so,

=>1/(cos²Asin²A) {since sin²A +cos²A =1}

=> 1/cos²A *1/sin²A

=>sec²A cosec²A

Which is equal to RHS

$<marquee behavior=move bgcolor= White > <h1> ☺️☺️Hence Proved✌️✌️ </h1></marquee>$

Answer 2

First taking LHS

sec2A +cosec2A

=1/cos2A + 1/sin2A

=(sin2A +cos²A)/(cos2Asin2A)

=1/(cos2Asin2A) {because sin2A +cos2A =1}

= 1/cos2A *1/sin2A

=sec2A cosec2A which is equal to RHS

hence proof .Mark it brainliest