If, tan theta = b/a then (acos theta + bsin theta)/(acos theta - bsin theta) = ?
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If, tan theta = b/a then (acos theta + bsin theta)/(acos theta - bsin theta) = ?
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Answer:
[tex]\boxed{\bf\: \dfrac{acos\theta + bsin\theta}{acos\theta - bsin\theta} = \dfrac{ { {a}^{2} + {b}^{2} }}{ {a}^{2} - {b}^{2} } \: } \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf\: tan\theta = \dfrac{b}{a} \\ [/tex]
Now, Consider
[tex]\sf\: \dfrac{acos\theta + bsin\theta}{acos\theta - bsin\theta} \\ [/tex]
Divide numerator and denominator by [tex]\sf\:cos\theta [/tex], we get
[tex]\sf\: = \: \dfrac{ \dfrac{acos\theta + bsin\theta}{cos\theta}}{ \dfrac{acos\theta - bsin\theta}{cos\theta}} \\ [/tex]
[tex]\sf\: = \: \dfrac{\dfrac{acos\theta}{cos\theta} + \dfrac{bsin\theta}{cos\theta} }{\dfrac{acos\theta}{cos\theta} - \dfrac{bsin\theta}{cos\theta} } \\ [/tex]
[tex]\sf\: = \: \dfrac{a + btan\theta}{a - btan\theta} [/tex]
On substituting the value of [tex]\sf\:tan\theta [/tex] from equation (1), we get
[tex]\sf\: = \: \dfrac{a + b \times \dfrac{b}{a}}{a - b \times \dfrac{b}{a}} \\ [/tex]
[tex]\sf\: = \: \dfrac{a + \dfrac{ {b}^{2} }{a}}{a - \dfrac{ {b}^{2} }{a}} \\ [/tex]
[tex]\sf\: = \: \dfrac{ { {a}^{2} + {b}^{2} } }{ {a}^{2} - {b}^{2} } \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\: \dfrac{acos\theta + bsin\theta}{acos\theta - bsin\theta} = \dfrac{ { {a}^{2} + {b}^{2} } }{ {a}^{2} - {b}^{2} } \: } \\ [/tex]