Question

If secϴ= x+1/4x , then the value of secϴ + tanϴ is​

Answer 1

Given ,

• secϴ = $\bold{x+\frac{1}{4x}}$

To find:-

• secϴ + tanϴ

We know that ,

$» \: \: \bold{ 1 + tan²ϴ = sec²ϴ}$

$→ tan \theta \: = ± \sqrt{sec ^{2} \theta - 1 }$

$→ tan \theta \: = ± \sqrt{ {(x + \frac{1}{4x}) }^{2} - 1 }$

$→ tan \theta \: = ± \sqrt{ { (\frac{4 {x}^{2} + 1 }{4x} } )^{2} - 1 }$

$→ tan \theta \: = ± \sqrt{ { (\frac{(4 {x}^{2}) ^{2} +2 \times 4 {x}^{2} \times 1 + { 1 }^{2} }{16 {x}^{2} }) } - 1 }$

$→ tan \theta \: = ± \sqrt{ { (\frac{16 {x}^{4} +8{x}^{2} + 1 }{16 {x}^{2} }) } - 1 }$

$→ tan \theta \: = ± \sqrt{ { (\frac{16 {x}^{4} +8{x}^{2} + 1 - 16 {x}^{2} }{16 {x}^{2} }) } }$

$→ tan \theta \: = ± \sqrt{ { (\frac{16 {x}^{4} - 8{x}^{2} + 1 }{16 {x}^{2} }) } }$

$→ tan \theta \: = ± \sqrt{ { (\frac{ {(4 {x}^{2}) }^{2} - 2 \times 4{x}^{2} \times 1 + {1}^{2} }{ {(4 x )}^{2} }) } }$

$→ tan \theta \: = ± \sqrt{ { (\frac{ {(4 {x}^{2} - 1) }^{2} }{ {(4 x )}^{2} }) } }$

$→ tan \theta \: = ± \sqrt{ { (\frac{ 4 {x}^{2} - 1 }{ 4 x }) } ^{2} }$

$→ tan \theta \: = ± { (\frac{ 4 {x}^{2} - 1 }{ 4 x }) }$

$→ \bold{\underline{tan \theta \: = ± (x - \frac{1}{4x} )}}$

★ If tanϴ = $\bold{x\: – \frac{1}{4x}}$

Then , secϴ + tanϴ

= x + 1/4x + x – 1/4x

= 2x

★ If tanϴ = $\bold{–(x\: – \frac{1}{4x})}$

Then , secϴ + tanϴ

= x + 1/4x + {–(x – 1/4x)}

= x + 1/4x – (x – 1/4x)

= x + 1/4x – x + 1/4x

= 2 × 1/4x

= 1/2x