Math, asked by krishakhetwani, 8 months ago

sinA-cosA+1/sinA+cosA-1=1/secA-tanA​

Answers

Answered by mariashakeb123
1

Step-by-step explanation:

LHS = sinA-cosA+1/sinA+cosA-1

divide both numerator and denominator by cosA

LHS=(tanA−1+secA)/(tanA+1−secA)

Now

sec2A=1+tan2A

sec2A−tan2A=1

Using above relation at denominator of LHS

LHS=(tanA−1+secA)/(tanA−secA+sec2A−tan2A)

LHS=(tanA−1+secA)/((secA−tanA)(−1+secA+tanA))

LHS=1/(secA−tanA)

LHS=RHS

Hence Proved.

Answered by MysteriousAryan
0

Answer:

refer \: to \: the \: attachment

Attachments:
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