Question

if m= a secA + b tanA and n= a tanA + b secA then prove m^2 - n^2 = a^2 - b^2​

Answer 1

Answer:

The answer is given in the attachment

Hope it helps

Answer 2

✪ᴘʀᴏᴏғ✪

Here,

$m= a secA + btanA$

$m^2= a^2sec^2 A + 2absecAtanA + b^2tan^2A$

And

$n= a tanA + b secA$

$n^2 = a^2tan^2A + 2absecAtanA + b^2sec^2A$

Sunstracting, we get

$m^2 - n^2$

$= a^2(sec^2 A - tan^2A) + b^2(tan^2A-sec^2A)$

$= a^2(sec^2 A - tan^2A) - b^2(sec^2 A - tan^2A)$

We know that

$\to 1 + tan^2A = sec^2A$

$\to sec^2A - tan^2A = 1$

So,

$m^2 - n^2 = a^2 - b^2$

Hence Proved