Question

if tan teta=p/q,then find p sin teta-q cos teta by pain teta+q cos teta​

Answer 1

Please refer the attached solution

Simply take q common and replace p/q with tan∅ while tan∅ with sin∅/cos∅.

Formulas Used:

→ cos2∅ = cos²∅ - sin²∅

Answer 2

$\sf{Answer}$

Step-by-step-explanation :-

Appropriate Queation:-

tanθ = $\sf\dfrac{p}{q}$ Then find the value of

$\sf\dfrac{psinθ-qcosθ}{psinθ+qcosθ}$

Solution :-

tanθ = $\sf\dfrac{opposite}{adjacent}$

From pythagoras theoram

AC² = AB² + BC²

(Hyp)² = s² + s²

(Hyp)² = p² + q²

Hyp = √p²+q²

Now ,

sinθ = $\sf\dfrac{opposite}{hypotenuse}$

sinθ = $\sf\dfrac{p}{√p²+q²}$

cosθ = $\sf\dfrac{q}{√p²+q²}$

Plugging values !

$\sf\dfrac{psinθ-qcosθ}{psinθ+qcosθ}$

psinθ- qcosθ= $\sf\dfrac{p²/√p²+q²}{-q²/√p²+q²}$

psinθ- qcosθ= $\sf\dfrac{p²-q²}{√p²+q²}$

psinθ+qcosθ = $\sf\dfrac{p²/√p²+q²}{+q²/√p²+q²}$

psinθ+qcosθ = $\sf\dfrac{p²+q²}{√p²+q²}$

$\sf\dfrac{psinθ-qcosθ}{psinθ+qcosθ}$ = $\sf\dfrac{p²-q²}{p²+q²}$

alternate method:-

tanθ = $\sf\dfrac{p}{q}$

tanθ = $\sf\dfrac{sinθ}{cosθ}$

So,

$\sf\dfrac{sinθ}{cosθ}$ = $\sf\dfrac{p}{q}$

sinθ = p

cosθ = q

Substuite in question

$\sf\dfrac{psinθ-qcosθ}{psinθ+qcosθ}$ =

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

$\sf\dfrac{p²-q²}{p²+q²}$

So,

$\sf\dfrac{psinθ-qcosθ}{psinθ+qcosθ}$ = $\sf\dfrac{p²-q²}{p²+q²}$

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If u cant understand clearly once refer attachments of solution..