Question

If tan teeta = p/q then find p sin teeta – q cos teeta /
p sin teeta + q cos teeta .


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Answer 1

Given :

 $\bullet\ \; \sf \blue{tan\ \theta = \dfrac{p}{q}}$

To Find :

$\bullet\ \; \sf \red{\dfrac{ p\ sin\ \theta -q\ cos\ \theta }{p\ sin\ \theta +q\ cos\ \theta }}$

Solution :

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

By taking cos θ common , we get ,

$:\implies \sf \dfrac{cos\ \theta\ \bigg(p\ \frac{sin\ \theta}{cos\ \theta}- q\ \frac{cos\ \theta}{cos\ \theta} \bigg) }{ cos\ \theta\ \bigg(p\ \frac{sin\ \theta}{cos\ \theta} + q\ \frac{cos\ \theta}{cos\ \theta} \bigg) }$

$\bullet\ \; \sf \pink{tan\ \theta = \dfrac{sin\ \theta}{cos\ \theta}}$

$:\implies \sf \dfrac{p\ tan\ \theta - q}{p\ tan\ \theta +q}$

$\bullet\ \; \sf \orange{tan\ \theta =\dfrac{p}{q}}\ \; [\ \because\ Given\ ]$

$:\implies \sf \dfrac{ p\ \bigg(\dfrac{p}{q} \bigg) - q }{ p\ \bigg( \dfrac{p}{q} \bigg) +q }$

$:\implies \sf \dfrac{ \dfrac{p^2}{q}-q }{\dfrac{ p^2}{q}+q }$

$:\implies \sf \dfrac{\dfrac{p^2-q^2}{q}}{\dfrac{p^2+q^2}{q} }$

$:\implies \sf \green{\dfrac{p^2-q^2}{p^2+q^2}}\ \; \bigstar$

$\bullet\ \; \sf \dfrac{ p\ sin\ \theta-q\ cos\ \theta }{ p\ sin\ \theta+q\ cos\ \theta }= \dfrac{p^2-q^2}{p^2+q^2}$