Question

27 can anyone help??

Answer 1

Trigonometry,

We have,
(secA+tanA)(secB+tanB)(secC+tanC) = (secA-tanA)(secB-tanB)(secC-tanC)

Now,
Let this equal to k
So (secA+tanA)(secB+tanB)(secC+tanC) = (secA-tanA)(secB-tanB)(secC-tanC) = k

Therefore k²
= k×k
(secA+tanA)(secB+tanB)(secC+tanC) × (secA-tanA)(secB-tanB)(secC-tanC) [as k is equal to both of them]
= (sec²A-tan²A)(sec²B-tan²B)(sec²C-tan²C)
= 1×1×1 [as sec²A - tan²A = 1]
= 1

So k = √1 = ±1

or, (secA+tanA)(secB+tanB)(secC+tanC) = (secA-tanA)(secB-tanB)(secC-tanC) = ±1 [proved]

That's it
Hope it helped ლ(´ڡ`ლ)