Math, asked by WalkZz, 11 months ago

sinA-cosA+1/sinA+cosA-1=1+sinA/cosA​

Answers

Answered by sandy1816
1

Step-by-step explanation:

sinA-cosA+1/sinA+cosA-1

devide cosA in both numerator & denominator

tanA-1+secA/tanA+1-secA

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

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

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

=tanA+secA

=sinA/cosA+1/cosA

=sinA+1/cosA

Answered by rishu6845
1

Step-by-step explanation:

formula

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a^2-b^2=(a+b)(a-b)

Attachments:
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