Question

If sec A = `sqrt2`, find the value of :
`(3cos^2"A"+5tan^2"A")/(4tan^4"A"–sin^2"A")`

Answer 1

Consider the diagram below :

sec A =`sqrt2`

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

Therefore if length of AB = x, length of AC =`sqrt2` 

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

(x)² + BC² = (`sqrt2"x")^2`

BC² = 2x² – x² = x²

∴ BC = x ...(perpendicular)

Now

tan A = `"perpendicular"/"base" = ("x")/("x") =1`

sin A = `"perpendicular"/"hypotenuse" = ("x")/(sqrt2"x") = (1)/(sqrt2)`

cos A = `"base"/"hypotenuse" = ("x")/(sqrt2"x") = (1)/(sqrt2)`

Therefore
`(3cos^2"A"+5tan^2"A")/(4tan^4"A"–sin^2"A")`

= `(3(1/sqrt2)^2+5(1)^2)/(4(1)^2 – (1/sqrt2)^2)`

= `(13/2)/(7/2)`

= `(13)/(7)`

=`1(6)/(7)`