Question

If 4 = sec and = tan, then find the value of 16(x^2-1/x^2)​

Answer 1

Answer:

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Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{4x = \sec \theta \: \: \: and \: \: \: \frac{4}{x} = \tan \theta}$

TO DETERMINE

$\displaystyle \sf{16 \bigg( {x}^{2} - \frac{1}{ {x}^{2} } \bigg) }$

EVALUATION

Here it is given that

$\displaystyle \sf{4x = \sec \theta \: \: \: and \: \: \: \frac{4}{x} = \tan \theta}$

Now

$\displaystyle \sf{16 \bigg( {x}^{2} - \frac{1}{ {x}^{2} } \bigg) }$

$\displaystyle \sf{ =\bigg( 16{x}^{2} - \frac{16}{ {x}^{2} } \bigg) }$

$\displaystyle \sf{ ={(4x)}^{2} - { \bigg( \frac{4}{x} \bigg)}^{2} }$

$\displaystyle \sf{ ={ \sec}^{2} \theta- { \tan}^{2} \theta }$

$= 1$

FINAL ANSWER

$\displaystyle \sf{16 \bigg( {x}^{2} - \frac{1}{ {x}^{2} } \bigg) = 1 }$

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