Question

Please help me in question no.13

Answer 1

Solution: (Instead of θ, I use A)

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Given:

sec A =$x + \frac{1}{4x}$

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To Prove:

sec A + tan A = 2x (or) $\frac{1}{2x}$

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Proof:

We know that,

sec² A - tan² A = 1

=> sec²A = 1+ tan²A

=> sec²A - 1 = tan²A

=> $\sqrt{sec^2A- 1 } = tan A$ ..(ii)

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Substituting value of x in (i),

We get,

=>$\sqrt{(x + \frac{1}{4x})^2 - 1 } = tanA$

=> $\sqrt{x^2 + (2)(\frac{1}{4x})(x) + ( \frac{1}{4x})^2 - 1 } = tan A$

=> $\sqrt{x^2 + \frac{2x}{4x} + ( \frac{1}{4x})^2 - 1 } = tanA$

=> $\sqrt{x^2 + \frac{1}{2} + ( \frac{1}{4x} )^2 -1 } = tanA$

=> $\sqrt{x^2 - \frac{1}{2} + ( \frac{1}{4x} )^2} = tanA$

=> $\sqrt{(x)^2 - (2)( \frac{1}{4x} )(x)+ ( \frac{1}{4x} )^2} = tanA$

=> $\sqrt{(x - \frac{1}{4x} )^2} = tan A$

  => ∴ $x - \frac{1}{4x} = tan A$

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secA + tan A

=> $(x + \frac{1}{4x}) + (x - \frac{1}{4x} )$

=> $2x + \frac{1}{4x} - \frac{`1}{4x}$

=> 2x,.

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As we know,

sec²A - tan²A = 1

As, we know that anything x  it's reciprocal is 1,

(secA + tan A) x $\frac{1}{secA + tanA}$ = 1

and hence, it has another possiblity,

=> $\frac{1}{secA + tanA}$ = sec A - tan A .,.(another possiblity),.

=> $\frac{1}{2x}$

∴ sec A + tan A = $\frac{1}{2x}$

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                                  Hope it Helps !!

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