Question

If Q tan theta = p, then find the value of Psin theta - Qcos theta / Psin theta + Q cos theta​

Answer 1

Answer:

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \sf{ \tan \theta = \frac{p}{q} }$

TO DETERMINE

$\displaystyle \sf{ \frac{p \sin \theta - q \cos \theta}{ p \sin \theta + q \cos \theta} }$

CALCULATION

$\displaystyle \sf{ \frac{p \sin \theta - q \cos \theta}{ p \sin \theta + q \cos \theta} }$

$\sf{Dividing \: numerator \: and \: denominator \: both \: by \: \cos \theta }$

= $\displaystyle \sf{ \frac{p \: \frac{\sin \theta}{\cos \theta} - q\: \frac{\cos \theta}{\cos \theta} }{ p \: \frac{\sin \theta}{\cos \theta} + q\: \frac{\cos \theta}{\cos \theta} } }$

= $\displaystyle \sf{ \frac{p \tan \theta - q }{ p \tan \theta + q } }$

= $\displaystyle \sf{ \frac{p \times \frac{p}{q} - q }{ p \times \frac{p}{q} + q } }$

= $\displaystyle \sf{ \frac{ \frac{ {p}^{2} - {q}^{2} }{q} }{ \frac{ {p}^{2} + {q}^{2} }{q} }}$

= $\displaystyle \sf{ \frac{ { {p}^{2} - {q}^{2} } }{ {p}^{2} + {q}^{2} }}$

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Step-by-step explanation:

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Answer 2

Answer:

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \sf{ \tan \theta = \frac{p}{q} }$

TO DETERMINE

$\displaystyle \sf{ \frac{p \sin \theta - q \cos \theta}{ p \sin \theta + q \cos \theta} }$

CALCULATION

$\displaystyle \sf{ \frac{p \sin \theta - q \cos \theta}{ p \sin \theta + q \cos \theta} }$

$\sf{Dividing \: numerator \: and \: denominator \: both \: by \: \cos \theta }$

= $\displaystyle \sf{ \frac{p \: \frac{\sin \theta}{\cos \theta} - q\: \frac{\cos \theta}{\cos \theta} }{ p \: \frac{\sin \theta}{\cos \theta} + q\: \frac{\cos \theta}{\cos \theta} } }$

= $\displaystyle \sf{ \frac{p \tan \theta - q }{ p \tan \theta + q } }$

= $\displaystyle \sf{ \frac{p \times \frac{p}{q} - q }{ p \times \frac{p}{q} + q } }$

= $\displaystyle \sf{ \frac{ \frac{ {p}^{2} - {q}^{2} }{q} }{ \frac{ {p}^{2} + {q}^{2} }{q} }}$

= $\displaystyle \sf{ \frac{ { {p}^{2} - {q}^{2} } }{ {p}^{2} + {q}^{2} }}$

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