Math, asked by bindut948, 11 months ago

SinA×cosA (sec squareA+cosec square A)= secA×tanA

Answers

Answered by sandy1816

1

Step-by-step explanation:

sinAcosA(sec²A+cosec²A)

=sinAcosA(1/cos²A+1/sin²A)

=sinAcosA(sin²A+cos²A/sin²Acos²A)

=sinAcosA/sin²Acos²A

=1/sinAcosA

=secAcosecA

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