Question

please no fake anwers Or other than topi.c..​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{p \:sin \: \theta - q \: cos \: \: \theta}{p \: sin \: \theta + q \: cos \: \theta} = \frac{ {p}^{2} - {q}^{2} }{ {p}^{2} + {q}^{2} }}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline\bold{Given :}} \\ \tt: \implies tan \: \theta = \frac{p}{q} \\ \\ \red{\underline\bold{To \:Find:}} \\ \tt: \implies \frac{p \:sin \: \theta - q \: cos \: \: \theta}{p \: sin \: \theta + q \: cos \: \theta} =$

• According to given question :

$\tt \circ \: tan \: \theta = \frac{p}{q} = \frac{p}{b} \\ \\ \tt \circ \: Perpendicular = p \\ \\ \tt \circ \: Base = q \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \tt: \implies {h}^{2} = {p}^{2} + {q}^{2} \\ \\ \tt: \implies h = \sqrt{ {p}^{2} + {q}^{2} } \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies \frac{p \:sin \: \theta - q \: cos \: \: \theta}{p \: sin \: \theta + q \: cos \: \theta} \\ \\ \tt: \implies \huge{\frac{p \times \frac{p}{h} - q \times \frac{b}{h} }{p \times \frac{p}{h} +q \times \frac{b}{h} }} \\ \\ \tt: \implies \huge{\frac{ \frac{p \times p}{ \sqrt{ {p}^{2} + {q}^{2} } } - \frac{q \times q}{ \sqrt{ {p}^{2} + {q}^{2} } } }{ \frac{p \times p}{ \sqrt{ {p}^{2} + {q}^{2} } }+ \frac{q \times q}{ \sqrt{{p}^{2} + {q}^{2} }} } }\\ \\ \tt: \implies \huge{\frac{ \frac{ {p}^{2} - {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } }{ \frac{ {p}^{2} + {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } } } \\ \\ \tt: \implies \frac{ {p}^{2} - {q}^{2} }{ {p}^{2} + {q}^{2} } \times \frac{ \sqrt{ {p}^{2} + {q}^{2} } }{ \sqrt{ {p}^{2} + {q}^{2} } } \\ \\ \tt: \implies \frac{ {p}^{2} - {q}^{2} }{ {p}^{2} + {q}^{2} } \\ \\ \green{\tt \therefore \frac{p \:sin \: \theta - q \: cos \: \: \theta}{p \: sin \: \theta + q \: cos \: \theta} = \frac{ {p}^{2} - {q}^{2} }{ {p}^{2} + {q}^{2} } }$

Answer 2

Aɴꜱᴡᴇʀ

➜ Option C

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Gɪᴠᴇɴ

$\tt tan \: \theta = \frac{p}{q}$

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ᴛᴏ ꜰɪɴᴅ

$\tt \large\frac{p \:sin \: \theta - q \: cos \: \: \theta}{p \: sin \: \theta + q \: cos \: \theta}$

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Sᴛᴇᴘꜱ

☞ As per the question

$\begin{lgathered}\tt \: tan \: \theta = \frac{p}{q} = \frac{p}{b} \\ \\ \tt \: Perpendicular = p \\ \\ \tt \: Base = q \\ \\ \bold{with \: the \: use \: of} \\ \\ \tt \leadsto {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \tt \leadsto {h}^{2} = {p}^{2} + {q}^{2} \\ \\ \tt \leadsto h = \sqrt{ {p}^{2} + {q}^{2} } \\ \\ \bold{now \: fining \: the \:values } \\ \\ \tt \dashrightarrow\frac{p \:sin \: \theta - q \: cos \: \: \theta}{p \: sin \: \theta + q \: cos \: \theta} \\ \\ \tt\dashrightarrow{\frac{p \times \frac{p}{h} - q \times \frac{b}{h} }{p \times \frac{p}{h} +q \times \frac{b}{h} }} \\ \\ \tt \dashrightarrow{\frac{ \frac{p \times p}{ \sqrt{ {p}^{2} + {q}^{2} } } - \frac{q \times q}{ \sqrt{ {p}^{2} + {q}^{2} } } }{ \frac{p \times p}{ \sqrt{ {p}^{2} + {q}^{2} } }+ \frac{q \times q}{ \sqrt{{p}^{2} + {q}^{2} }} } }\\ \\ \tt\dashrightarrow{\frac{ \frac{ {p}^{2} - {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } }{ \frac{ {p}^{2} + {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } } } \\ \\ \tt \dashrightarrow \frac{ {p}^{2} - {q}^{2} }{ {p}^{2} + {q}^{2} } \times \frac{ \sqrt{ {p}^{2} + {q}^{2} } }{ \sqrt{ {p}^{2} + {q}^{2} } } \\ \\ \tt \dashrightarrow \frac{ {p}^{2} - {q}^{2} }{ {p}^{2} + {q}^{2} } \\ \\ \pink{\tt \dashrightarrow \frac{p \:sin \: \theta - q \: cos \: \: \theta}{p \: sin \: \theta + q \: cos \: \theta} = \frac{ {p}^{2} - {q}^{2} }{ {p}^{2} + {q}^{2} } }\end{lgathered}$

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