Question

RS Aggarwal Exercise 10 , Question no.7​

Answer 1

$\sf \sin\theta \: = \dfrac{c}{ \sqrt{ {c}^{2} + {d}^{2} } } \\ \\ \gray{\sf {we \: know \: the \: formula,}} \\ \\ \boxed{ \gray{\underline{ \sf{ {cos}^{2} \theta = 1 - {sin}^{2} \theta }}}} \\ \\ \sf \: {cos}^{2} \theta = 1 - ( \frac{c}{ \sqrt{ {c}^{2} + {d}^{2} } } ) ^{2} \\ \\ \sf \: {cos}^{2} \theta = 1 - \frac{ {c}^{2} }{ {c}^{2} + {d}^{2} } \\ \\ \sf \: {cos}^{2} \theta = \frac{ \cancel{{c}^{2} }+ {d}^{2} - \cancel{{c}^{2} }}{ { {c}^{2} + {d}^{2} }} \\ \\ \sf {cos} ^{2} \theta = \frac{ {d}^{2} }{ {c}^{2} + {d}^{2} } \\ \\ \sf \: cos \theta = \sqrt{ \frac{ {d}^{2} }{ {c}^{2} + {d}^{2} } } \\ \\ \boxed{ \red{ \sf \: cos\theta = \frac{d}{ \sqrt{ {c}^{2} + {d}^{2} } } }} \\ \\ \\ \gray{ \sf{we \: know \: the \: formula,}} \\ \boxed{ \gray{ \underline{ \sf{tan\theta = \frac{sin\theta }{cos\theta } }}}} \\ \\ \sf \: tan\theta = \dfrac{ \dfrac{c}{ \sqrt{ {c}^{2} + {d}^{2} } } }{ \dfrac{d}{ \sqrt{ {c}^{2} + {d}^{2} } } } \\ \\ \sf \: tan\theta = \dfrac{ \dfrac{c}{ \cancel{\sqrt{ {c}^{2} + {d}^{2} } } } }{ \dfrac{d}{ \cancel{ \sqrt{ {c}^{2} + {d}^{2} } } }} \\ \\ \boxed{ \red{\sf \: tan \theta = \frac{c}{d} }}$