Question

please mark these answers fast friends
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Answer 1

Step-by-step explanation:

$\large \rm\pink {\underline{ { \blue{2. } \: \: \green {Solution :-}}}}$

$\rm{ \cot \theta = \frac{base}{height} = \frac{q}{p} }$

From Pythagoras theorem:-

$\rm (hypotenuse)^2 = (height)^2 + (base)^2$

$\rm (hypotenuse)^2 = p^2 + q^2$

$\rm hypotenuse = \sqrt{p^2 + q^2}$

$\rm sin \theta = \frac{height }{hypotenuse} = \frac{p}{ \sqrt{ {p}^{2} + {q}^{2} } }$

$\rm cos\theta = \frac{base }{hypotenuse} = \frac{q}{ \sqrt{ {p}^{2} + {q}^{2} } }$

$\rm LHS = \frac{psin \theta \: - \: qcos \theta}{psin \theta \: + \: qcos \theta}$

$\rm \: \: = \frac{p .\frac{p}{ \sqrt{ {p}^{2} + {q}^{2} } } \: - \: q. \frac{q}{ \sqrt{ {p}^{2} + {q}^{2} } } }{p .\frac{p}{ \sqrt{ {p}^{2} + {q}^{2} } } \: + \: q. \frac{q}{ \sqrt{ {p}^{2} + {q}^{2} } } }$

$\rm \: \: = \frac{\frac{ {p}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } \: - \: \frac{ {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } }{\frac{ {p}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } \: + \: \frac{ {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } }$

$\rm \: \: = \frac{\frac{ {p}^{2} - {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } }{ }{ } }{\frac{ {p}^{2} + {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } }}$

$\rm \: \: = {\frac{ {p}^{2} - {q}^{2} }{ \cancel{\sqrt{ {p}^{2} + {q}^{2}}}}} \times { \frac{ \: \: \cancel{\sqrt{ {p}^{2} + {q}^{2} } } }{ {p}^{2} + {q}^{2}}}$

$\rm = \frac{ {p}^{2} - {q}^{2} }{{p}^{2} + {q}^{2}}$

$\rm = RHS$

Answer 2

$\sf{ \cot \theta = \frac{base}{height} = \frac{q}{p} }$

From Pythagoras theorem:-

$\sf (hypotenuse)^2 = (height)^2 + (base)^2$

$\sf(hypotenuse)^2 = p^2 + q^2$

$\sf hypotenuse = \sqrt{p^2 + q^2}$

$\sf sin \theta = \frac{height }{hypotenuse} = \frac{p}{ \sqrt{ {p}^{2} + {q}^{2} } }$

$\sf cos\theta = \frac{base }{hypotenuse} = \frac{q}{ \sqrt{ {p}^{2} + {q}^{2} } }$

$\sf LHS = \frac{psin \theta \: - \: qcos \theta}{psin \theta \: + \: qcos \theta}$

$\sf\large \: \: = \frac{p .\frac{p}{ \sqrt{ {p}^{2} + {q}^{2} } } \: - \: q. \frac{q}{ \sqrt{ {p}^{2} + {q}^{2} } } }{p .\frac{p}{ \sqrt{ {p}^{2} + {q}^{2} } } \: + \: q. \frac{q}{ \sqrt{ {p}^{2} + {q}^{2} } } }$

$\sf\large \: \: = \frac{\frac{ {p}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } \: - \: \frac{ {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } }{\frac{ {p}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } \: + \: \frac{ {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } } }$

$\sf\large \: \: = \frac{\frac{ {p}^{2} - {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } }{ }{ } }{\frac{ {p}^{2} + {q}^{2} }{ \sqrt{ {p}^{2} + {q}^{2} } }}$

$\sf\large \: \: = {\frac{ {p}^{2} - {q}^{2} }{ \cancel{\sqrt{ {p}^{2} + {q}^{2}}}}} \times { \frac{ \: \: \cancel{\sqrt{ {p}^{2} + {q}^{2} } } }{ {p}^{2} + {q}^{2}}}$

$\sf\large = \frac{ {p}^{2} - {q}^{2} }{{p}^{2} + {q}^{2}}$