Question

prove 1+secA - tanA/ 1+secA+tanA = 1-sinA/cosA​

Answer 1

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Answer 2

L.H.S = 1 + secA – tanA /1+ secA + tanA 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+b23(a+b) (a-b)]

L.H.S = (secA - tanA) (1+secA + tanA) / 1+ secA + tanA

L.H.S = secA – tanA

R.H.S=1

We know about the formula of secA and tanA, [secA = 1/ COSA] [tanA = sinA / cosA]

putting the value of secA and tanA= 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 LHS is equal to R.H.S