Question

Pls solve this answer fast.
Pls don’t give swamp answers .

Answer 1

$\huge\red{\frak{\underline{\underline{\:ANSWER}}}}$

$\sf{\pink{\underline{\:Given}}}$

$\green{\mapsto{\sf{\:\sec \theta \:=\:x+\dfrac{1}{4x}}}}$.....(1)

$\sf{\orange{\underline{\:Evaluate}}}$

$\green{\mapsto{\sf{\:(\sec \theta + \tan \theta)}}}$

$\LARGE\pink{\frak{\underline{\underline{\:EXPLANATION}}}}$

$\:\:\:\:\green{\sf{\:(We\:know)}}$

$\bold{\red{\boxed{\boxed{\pink{\:\tan^2 \theta \:=\sec^2 \theta \:-\:1}}}}}$

$\:\:\:\:\green{\sf{\:keep\:value\:by\:equ(1)}}$

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

$\:\:\:\:\:\green{\sf{\:(a+b)^2\:=\:(a^2+b^2+2ab)}}$

$\mapsto\sf{\:\tan^2 \theta\:=\:\left(x^2+\dfrac{1}{16x^2} + \dfrac{1}{2} - 1\right)}$

$\mapsto\sf{\:\tan^2 \theta\:=\:\left(x^2+\dfrac{1}{16x^2} - \dfrac{1}{2}\right)}$

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

$\mapsto\sf{\:\tan \theta\:=\:\sqrt{\left(x-\dfrac{1}{4x}\right)^2}}$

$\mapsto\sf{\:\tan \theta\:=\:±\left(x-\dfrac{1}{4x}\right)}$

Take ( + ) Sign

$\mapsto\sf{\:\tan \theta\:=\:\left(x-\dfrac{1}{4x}\right)}$

Take ( - ) Sign

$\mapsto\sf{\:\tan \theta\:=\:\left(-x+\dfrac{1}{4x}\right)}$

Case(1):-

• When,

$\red{\sf{\:\sec \theta\:=\:(x+\frac{1}{4x})\:\:and\:\:\tan \theta\:=\:(x-\frac{1}{4x})}}$

Value:-

$\pink{\sf{\:(\sec \theta + \tan \theta)}}$

$\mapsto\sf{\:(x+\:\cancel{\dfrac{1}{4x}})\:+\:(x-\cancel{\dfrac{1}{4x}})}$

$\mapsto\pink{\sf{\:(2x)}}$

Case(2):-

• When

$\red{\sf{\:\sec \theta\:=\:(x+\frac{1}{4x}\:\:and\:\:\tan \theta\:=\:(-x+\frac{1}{4x}}}$

Value

$\sf{\:(\sec \theta + \tan \theta)}$

$\mapsto\sf{\:(\cancel{x}+\:\dfrac{1}{4x})\:+\:(\cancel{-x}+\dfrac{1}{4x})}$

$\mapsto\pink{\sf{\:\left(\dfrac{1}{2x}\right)}}$

$\large\green{\underbrace{\:Thus}}$

$\:\:\:\red{\sf{\:Value\:Of\:(\sec \theta + \tan \theta)}}$

$\mapsto\pink{\sf{\:(2x)}}$

$\mapsto\pink{\sf{\:\left(\dfrac{1}{2x}\right)}}$

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