Question

if sec theta = x+1/4x,prove that sec theta+tan theta=2x​

Answer 1

Given

sec∅ = x + 1/4x

To prove

sec∅ + tan∅ = 2x

Proof

sec∅ = x + 1/4x

Squaring both sides,

→ sec²∅ = (x + 1/4x)²

→ sec²∅ = x² + 1/16x² + 1/2

Now, we know that

1 + tan²∅ = sec²∅

→ 1 + tan²∅ = x² + 1/16x² + 1/2 (since, sec²∅ = x² + 1/16x² + 1/2)

→ tan²∅ = x² + 1/16x² + 1/2 - 1

→ tan²∅ = x² + 1/16x² - 1/2

→ tan²∅ = (x - 1/4x)²

(using, a² + b² - 2ab = (a - b)²)

→ tan∅ = x - 1/4x

So, sec∅ + tan∅ = x + 1/4x + x - 1/4x

→ x + x

= 2x

Hence Proved.

Answer 2

Given :

• $\sf{sec\:\theta\:=\:x\:+\:\dfrac{1}{4x}}$

To prove :

• $\sf{sec\:\theta\:+\:tan\:\theta\:=\:2x}$

Proof :

$\sf{sec\:\theta\:=\:x\:+\:\dfrac{1}{4x}}$

Squaring both sides,

$\sf{sec^2\:\theta\:=\:\:\Bigg(x\:+\:\dfrac{1}{4x}\Bigg)^2}$

$\sf{sec^2\:\theta\:=\:x^2\:+\:2\:\times\:x\:\times\:\dfrac{1}{4x}\:+\:\big(\dfrac{1}{4}x\big)^2}$

$\sf{sec^2\:\theta\:=\:x^2\:+\:\dfrac{2}{4}\:+\:\dfrac{1}{16x^2}}$

$\sf{sec^2\:\theta\:=\:x^2\:+\:\dfrac{1}{2}\:+\:\dfrac{1}{16x^2}}$

We know,

• $\boxed{\blue{\sf{1\:+\:tan^2\:\theta\:=\:sec^2\:\theta}}}$

$\sf{1+tan^2\theta\:=\:x^2\:+\:\dfrac{1}{2}\:+\:\dfrac{1}{16x^2}}$

$\sf{tan^2\:\theta\:=\:x^2\:+\:\dfrac{1}{2}\:+\:\dfrac{1}{16x^2}\:-\:1}$

$\sf{tan^2\:\theta\:=\:x^2\:+\dfrac{1}{16x^2}\:+\:\dfrac{1}{2}-1}$

$\sf{tan^2\:\theta\:=\:x^2\:+\:\dfrac{1}{16x^2}\:+\:\dfrac{1-2}{2}}$

$\sf{tan^2\:\theta\:=\:x^2\:+\:\dfrac{1}{16x^2}\:-\:\dfrac{1}{2}}$

$\sf{tan^2\:\theta\:=\:\Big(x^2\:-\:\dfrac{1}{4x^2}\Big)}$

$\sf{tan\:\theta\:=\:\sqrt{x^2\:-\:\dfrac{1}{4x^2}}}$

$\sf{tan\:\theta\:=\:x\:-\:\dfrac{1}{4x}}$

$\sf{\red{Sec\:\theta\:=\:x\:+\:\dfrac{1}{4x}}}$

$\sf{\purple{Tan\:\theta\:=\:x\:-\:\dfrac{1}{4x}}}$

$\sf{sec\:\theta\:+\:tan\:\theta\:=\:x\:+\:\cancel\dfrac{1}{4x}}\:+\:x\:\:-\cancel\dfrac{1}{4x}$

$\sf{sec\:\theta\:+\:tan\:\theta\:=\:x\:+\:x}$

$\sf{sec\:\theta\:+\:tan\:\theta\:=\:2x}$

Hence proved.