Math, asked by jsoni2626, 11 months ago

15.
1+cos A+sin A
1+cos A-sin A
1+sin A
cos A​

Answers

Answered by sandy1816
1

Step-by-step explanation:

1+cosA+sinA/1+cosA-sinA

devide cosA in both numerator & denominator

=secA+1+tanA/secA+1-tanA

=(secA+tanA+1)(secA-tanA)/ (secA-tanA+1)(secA-tanA)

=sec²A-tan²A+secA-tanA/ (secA-tanA+1)(secA-tanA)

=1+secA-tanA/(secA-tanA+1)(secA-tanA)

=1/secA-tanA

=secA+tanA/sec²A-tan²A

=secA+tanA

=1/cosA+sinA/cosA

=1+sinA/cosA

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