Question

ANSWER THE FOLLOWING QUESTION IN STEPS:

Answer 1

Answer:

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