Question

1-secA+tanA/1+secA-tanA=secA+tanA-1/secA+tanA+1

Answer 1

Answer:

to solve this let us first understand the properties :

[secA]^2 – [tanA]^2 = 1

and

a^2 - b^2 = (a+b)(a-b)

Now take the numerator of LHS:

1 - sec A - tan A

instead of 1 , I can write , [secA]^2 – [tanA]^2

So the numerator now becomes : [secA]^2 – [tanA]^2 - (sec A + tan A)

which I can write : (sec A + tan A)(sec A - tan A) - (sec A - tan A)

taking (sec A - tan A) common,

numerator becomes , (sec A - tan A) (sec A + tan A -1)

using the same way , for denominator, by writing [secA]^2 – [tanA]^2 instead of 1 , and then writing it in the form (sec A + tan A)(sec A - tan A) and then taking (sec A - tan A) common,

denominator becomes , (sec A - tan A) (sec A + tan A +1)

Now writing numerator and denominator together : [(sec A - tan A) (sec A + tan A -1) ] / [(sec A - tan A) (sec A + tan A +1) ]

= (secA+tanA-1) / (secA+tanA+1)

= RHS

Hence Proved !

Step-by-step explanation:

mark brainliest

Answer 2

1-secA+tanA/1+secA-tanA=secA+tanA-1/secA+tanA+1

Answer : RHS