Question

if tan theta equals to p/q find the value of p sin - q sin /p sin + q sin ​

Answer 1

Answer:

Given     $tan x =\frac{p}{q}$

We need to find the value of

$\frac{psinx-qsinx}{psinx+qsinx}$

Step-by-step explanation:

$tan x =\frac{p}{q}$

Use pythagoras theorem ,

$p^{2} +q^{2} =x^{2} \\\\=> x = \sqrt{p^{2}+q^{2} }$

So $sinx=\frac{p}{\sqrt{p^{2}+q^{2} } }$

$psinx=\frac{p^{2} }{\sqrt{p^{2}+q^{2} } }\\\\qsinx=\frac{pq }{\sqrt{p^{2}+q^{2} } }$

So $psinx-qsinx = \frac{p^{2} }{\sqrt{p^{2}+q^{2} } }-\frac{pq }{\sqrt{p^{2}+q^{2} } }\\\\=> psinx-qsinx=\frac{p(p-q)}{\sqrt{p^{2}+q^{2} } } \\\\psinx+qsinx = \frac{p^{2} }{\sqrt{p^{2}+q^{2} } }+\frac{pq }{\sqrt{p^{2}+q^{2} } }\\\\=> psinx+qsinx=\frac{p(p+q)}{\sqrt{p^{2}+q^{2} } }$

Therefore $\frac{psinx-qsinx}{psinx+qsinx} = \frac{\frac{p(p-q)}{\sqrt{p^{2}+q^{2} } } )}{\frac{p(p+q)}{\sqrt{p^{2}+q^{2} } } )} \\\\=>\frac{psinx-qsinx}{psinx+qsinx}=\frac{(p-q)}{(p+q)}$

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