Question

If tanQ= p/q find the values of (p sinQ -q cosQ) / (psinQ+ qcosQ)

Answer 1

Step-by-step explanation:

We have,

$\rm \tan(Q) = \frac{p}{q}$

Now, we have,

$\rm \frac{p\sin(Q) - q \cos(Q) }{ p\sin(Q) + q \cos(Q) }$

Dividing num. and deno. by $\rm\:\cos(Q)$

So,

$\rm \implies \frac{p\sin(Q) - q \cos(Q) }{ p\sin(Q) + q \cos(Q) } = \frac{ \frac{p\sin(Q) - q \cos(Q) }{ \cos(Q) } }{ \frac{ p\sin(Q) + q \cos(Q) }{ \cos(Q) } }$

$\rm \implies \frac{p\sin(Q) - q \cos(Q) }{ p\sin(Q) + q \cos(Q) } = \frac{ \frac{p\sin(Q)}{ \cos(Q) } - \frac{ q \cos(Q) }{ \cos(Q) } }{ \frac{ p\sin(Q)}{ \cos(Q) } + \frac{ q \cos(Q) }{ \cos(Q) } }$

$\rm \implies \frac{p\sin(Q) - q \cos(Q) }{ p\sin(Q) + q \cos(Q) } = \frac{ p\tan(Q) - q }{ p\tan(Q) + q }$

$\rm \implies \frac{p\sin(Q) - q \cos(Q) }{ p\sin(Q) + q \cos(Q) } = \frac{ p \times \frac{p}{q} - q }{ p \times \frac{p}{q} + q }$

$\rm \implies \frac{p\sin(Q) - q \cos(Q) }{ p\sin(Q) + q \cos(Q) } = \frac{ \frac{p^{2} }{q} - q }{ \frac{p ^{2} }{q} + q }$

$\rm \implies \frac{p\sin(Q) - q \cos(Q) }{ p\sin(Q) + q \cos(Q) } = \frac{ \frac{p^{2} - {q}^{2} }{q} }{ \frac{p ^{2} + {q}^{2} }{q} }$

$\rm \implies \frac{p\sin(Q) - q \cos(Q) }{ p\sin(Q) + q \cos(Q) } = \frac{ p^{2} - {q}^{2} }{ p ^{2} + {q}^{2} }$

Answer 2

$\large\underline{\sf{Given- }}$

$\red{\rm :\longmapsto\:tanQ = \dfrac{p}{q}}$

$\large\underline{\sf{To\:Find - }}$

$\boxed{ \rm{ \frac{psinQ - qcosQ}{psinQ + qcosQ}}}$

$\large\underline{\sf{Solution-}}$

Given that,

$\red{\rm :\longmapsto\:tanQ = \dfrac{p}{q}}$

$\rm :\longmapsto\:\dfrac{sinQ}{cosQ} = \dfrac{p}{q}$

can be rewritten as by multiplying both sides by p/q.

$\rm :\longmapsto\:\dfrac{psinQ}{qcosQ} = \dfrac{ {p}^{2} }{ {q}^{2} }$

Now, apply Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{psinQ + qcosQ}{psinQ - qcosQ} = \dfrac{ {p}^{2} + {q}^{2} }{ {p}^{2} - {q}^{2} }$

Now, by applying invertendo, we get

$\rm :\longmapsto\:\dfrac{psinQ - qcosQ}{psinQ + qcosQ} = \dfrac{ {p}^{2} - {q}^{2} }{ {p}^{2} + {q}^{2} }$

Result Used :-

$\red{\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} , \: then}$

$\boxed{ \rm{ \frac{a + b}{b} = \frac{c + d}{d} \: is \: called \: componendo}}$

$\boxed{ \rm{ \frac{a - b}{b} = \frac{c - d}{d} \: is \: called \: dividendo}}$

$\boxed{ \rm{ \frac{a + b}{a - b} = \frac{c + d}{c - d} \: is \: called \: componendo \: and \: dividendo}}$

$\boxed{ \rm{ \dfrac{b}{a} = \dfrac{d}{c} \: is \: called \: invertendo}}$

Additional Information :-

$\boxed{ \rm{ {sin}^{2}x + {cos}^{2}x = 1}}$

$\boxed{ \rm{ {sec}^{2}x - {tan}^{2}x = 1}}$

$\boxed{ \rm{ {cosec}^{2}x - {cot}^{2}x = 1}}$