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Q) IF sec θ=(x + 1/4x) then prove that sec θ + tan θ = 2x or 1/2x

Ans:  

sec~\theta=x+\frac{1}{4x}

tan^{2}\theta=sec^{2}\theta-1

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

=x^{2}+\frac{1}{16x^{2}}+\frac{1}{2}-1

=x^{2}+\frac{1}{16x^{2}}-\frac{1}{2}

=(x-\frac{1}{4x})^{2}

\tan \theta = \pm \left(x - \frac{1}{4x}\right)

\implies tan\theta=(x-\frac{1}{4x})

or

tan\theta=-(x-\frac{1}{4x})

Given sec~\theta=x+\frac{1}{4x}

When

tan\theta=(x-\frac{1}{4x})

\therefore \sec \theta + \tan \theta = x + \cancel{\frac{1}{4x}} + x - \cancel{\frac{1}{4x}} = 2x

When

tan~\theta=-(x-\frac{1}{4x})

\therefore \sec \theta + \tan \theta = x + \frac{1}{4x} - (x - \frac{1}{4x})

= \cancel{x} + \frac{1}{4x} - \cancel{x} + \frac{1}{4x}

=\frac{1}{2x}

Therefore, value of \sec \theta + \tan \theta = 2 x~or~\frac{1}{2x}

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