Question

If sec ∅= x+1/4x, prove that :



Sec ∅+ tan∅= 2x or 1/2x







Regards,




Krishnachalilondon..♥️♥️

Answer 1

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Trigonometry

↪ concept is very clear what we are going to do we will find tan²@

↪ so later We can find tan@ to add with sec @

Now ,

we know from identity,

Tan²@ = Sec²@-1
=(x +1/4x)² -1
=x² +1/16x² + 2× x×1/4x -1
= (x-1/4x)²

Taking root both side

Tan @= x-1/4x

Now we have to find

Sec@+Tan@
=x+1/4x +x - 1/4x
=2x

HENCE proved....So simple

☺☺Bye Forever

Answer 2

$\huge{\boxed{\huge{\red\star\mathfrak\purple{\large{\underline{\underline{Answer!}}}}}}}$

We have,

$\huge \boxed{ \mathfrak \red{ \sec \theta = x + \frac{1}{4x} }}$

$\therefore \:$

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

$\bold \red{ \implies{ \tan \theta =( x + \frac{1}{4x} ) {}^{2 } - 1 }}$

$\bold \red{ \implies{ \tan \theta = x {}^{2} + \frac{1}{16x {}^{2} } + \frac{1}{2} - 1 }}$

$\bold \red{ \implies{ \tan \theta = (x - \frac{1}{4x} ) {}^{2} }}$

$\bold \red{ \implies \tan \theta = ± (x - \frac{1}{4x} )}$

$\bold \red{ \implies(x - \frac{1}{4x} ) \: or \: - ( x - \frac{1}{4x} )}$

When,

$\huge \boxed{ \mathfrak \pink{ \tan \theta = - (x - \frac{1}{4x} )}}$

$\bold \pink{\sec \theta + \tan \theta = x + \frac{1}{4x} + x - \frac{1}{4x} = 2x}$

When,

$\huge \boxed{ \mathfrak \purple{ \tan \theta = - (x - \frac{1}{4x} )}}$

$\bold \purple{ \sec \theta + \tan \theta = (x + \frac{1}{4x} ) - (x - \frac{1}{4x} )}$

$\bold \purple{ = \frac{2}{4x} = \frac{1}{2x} }$

Hence,

$\huge{{\huge{\boxed{\huge{\red\star\mathfrak\purple{\large{\underline{\underline{ \sec \theta + \tan \theta = 2x \: or \: \frac{1}{2x} }}}}}}}}}$

$\huge \boxed{ \mathfrak\orange{hence \: proved}}$