Question

If 3 tan A = 4 then prove that `sqrt((sec A - "cosec" A)/(sec A + "cosec" A)) = 1/sqrt(7)`.

Answer 1

Given: 3 tan A = 4

To prove: `sqrt((sec A - "cosec" A)/(sec A + "cosec" A)) = 1/sqrt(7)`

Proof:

`tan A = 4/3`

⇒ `(BC)/(AB) = 4/3`

Let BC = 4x and AB = 3x.

Then, AC² = AB² + BC²

= (9x² + 16x²)

= 25x²

⇒ `AC = sqrt(25x^2)`

⇒ AC = 5x

∴ `sin A = (BC)/(AC) = (4x)/(5x) = 4/5`

`cos A = (3x)/(5x) = 3/5`

`"cosec"  A = 1/(sin A) = 5/4`

And `sec A = 1/(cos A) = 5/3`

 `((sec A - "cosec" A))/((sec A + "cosec" A)) = ((5/3 - 5/4))/((5/3 + 5/4))`

= `((5/12))/((35/12))`

= `5/35`

= `1/7`

⇒ `sqrt((sec A - "cosec" A)/(sec A + "cosec" A)) = sqrt(1/7)`

= `1/sqrt(7)`