Question

if product of intercepts made by straight line xtan alpha + y sec alpha = 1 on the coordinaate axes is equal to sinalpha find alpha

Answer 1

SOLUTION

GIVEN

The product of intercepts made by straight line

$\sf{ x\tan \alpha + y \sec \alpha = 1 }$

on the coordinate axes is equal to $\sf{ \sin \alpha }$

TO DETERMINE

$\sf{ The \: value \: of \: \: \alpha }$

EVALUATION

The given equation of the line is

$\sf{ x\tan \alpha + y \sec \alpha = 1 }$

Which can be rewritten as

$\displaystyle \sf{ \frac{x}{ \cot \alpha } + \frac{y}{ \cos \alpha } = 1 }$

$\sf{Thus \: the \: line \: intersects \: x \: axis \: at \: \: ( \cot \alpha , 0)}$

$\sf{and \: \: x \: axis \: at \: \: ( 0,\cos \alpha )}$

So the product of intercepts

$= \sf{ \cot \alpha . \cos \alpha \: }$

$= \displaystyle \sf{ \frac{ { \cos}^{2} \alpha }{ { \sin} \alpha } \: }$

Now by the given condition

$\displaystyle \sf{ \frac{ { \cos}^{2} \alpha }{ { \sin} \alpha } = { \sin} \alpha \: }$

$\implies \displaystyle \sf{ { \cos}^{2} \alpha = {{ \sin}}^{2} \alpha }$

$\implies \displaystyle \sf{ { \cos} \alpha = \pm \: {{ \sin}} \alpha }$

$\implies \displaystyle \sf{ { \cot} \alpha = \pm 1 }$

$\implies \displaystyle \sf{ { \cot} \alpha = \pm \cot \frac{\pi}{4} }$

$\implies \displaystyle \sf{ \alpha = n\pi \pm \: \frac{\pi}{4} }$

$\displaystyle \sf{One \: of \: the \: possible \: values \: of \: \alpha \: is \: \: \frac{\pi}{4} }$

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