(secA tanA) (I-sin A ) =cosA
Answers
Answered by
1
Answer:
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
Mark it as branliest plzzzz
Similar questions