Question

If sec A = `sqrt2` , find : `(3cot^2 "A"+ 2 sin^2 "A")/ (tan^2 "A" – cos ^2 "A")`.

Answer 1

Consider the figure :

sec A = `sqrt2/1`

i.e.`"hypotenuse"/"base" = "AC"/"AB" = sqrt2/1`

Therefore if length of base = x , length of hypotenuse = `sqrt2x`

Since
AB² + BC² = AC²  ...[Using Pythagoras Theorem]

`(sqrt2x)^2 – (x)^2 = "BC"^2`

`"BC"^2 = 2x^2 - x^2`

BC² = x²

∴ BC = x

Now

cos A = `1/(sec "A") = 1/(sqrt2)`

sin A = `"BC"/"AC" = 1/(sqrt2)`

tan A = `"BC"/"AB"` = 1

cot A = `1/ tan "A"` = 1

Therefore

`(3cot^2 "A"+ 2 sin^2 "A")/ (tan^2 "A" – cos ^2 "A") = (3(1)^2 + 2 (1/sqrt2)^2)/ (1^2 – ( 1/sqrt2)^2)`

= `(3 + 1)/(1– (1)/(2)`

`= 4/(1/2)`

`= 4 xx 2/1`

= 8