Question

If `sec A = x + 1/(4x)`, then show that sec A + tan A = 2x or `1/(2x)`.

Answer 1

`sec A = x + 1/(4x)`   ...[Given]

We know that,

1 + tan²A = sec²A

∴ tan²A = sec²A – 1

= `(x + 1/(4x))^2 - 1`

= `x^2 + 2 xx x xx 1/(4x) + (1/(4x))^2 - 1`   ...[∵ (a + b)² = a² + 2ab + b²]

= `x^2 + 1/2 + 1/(16x^2) - 1`

= `x^2 - 1/2 + 1/(16x^2)`

∴ `tan^2A = (x - 1/(4x))^2`   ...[∵ a² – 2ab + b² = (a – b)²]

∴ `tan A = x - 1/(4x)` or `tan A = -(x - 1/(4x))`

When `tan A = x - 1/(4x)`,

sec A + tan A

= `x + 1/(4x) + x - 1/(4x)`

= 2x

When `tan A = -(x - 1/(4x))`,

sec A + tan A

= `x + 1/(4x) - (x - 1/(4x))`

= `x + 1/(4x) - x + 1/(4x)`

= `2/(4x)`

= `1/(2x)`