prove that 1+secA-tanA/1+secA+tanA=1-sinA/cosA
ij22:
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i got the correct answer
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L.H.S = 1 + secA – tanA / 1 + secA + tanA , R.H.S =1 – sinA / cosA
= L.H.S(sec2A – tan2A) + secA – tanA / 1 + secA + tanA
As we know that [sec2A – tan2A = 1]
So here L. H. S= (secA – tanA) (secA + tanA) + (secA – tanA) / 1 + secA + tanA
We know about this formula [a2+b2=(a+b) (a-b)]
L.H.S = (secA – tanA) (1+secA + tanA) / 1 + secA + tanA
L.H.S = secA – tanA
We know about the formula of secA and tanA,[secA = 1 / cosA], [ tanA = sinA / cosA]
putting the value of secA and tanA
so = 1 / cosA - sinA / cosA
L.H. S= 1 – sinA / cosA
So here L.H. S is equal to R.H.S
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