Question

find me the answer plz​

Answer 1

Answer:

the value of theta = 45°

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Answer 2

—› Solution :-

$\implies \tt \: 3 \: tan( \theta - {15}^{0}) = tan( \theta + {15}^{0} )$

$\implies \tt \: \frac{3}{1} = \frac{tan( \theta + {15}^{0}) }{tan( \theta - {15}^{0}) }$

We know that,

According to componendo and dividendo,

$\boxed{ \gray {\sf \: \frac{a}{b} = \frac{c}{d} \implies \frac{a + b}{a - b} = \frac{c + d}{c - d} } }$

$\implies \tt \: \frac{3 + 1}{3 - 1} = \frac{tan( \theta + {15}^{0} ) + tan( \theta - {15}^{0}) }{tan( \theta + {15}^{0} ) - tan( \theta - {15}^{0}) }$

$\implies \tt \: \frac{ \cancel4 \: {}^{2} }{ \cancel2} = \frac{tan( \theta + {15}^{0} ) + tan( \theta - {15}^{0}) }{tan( \theta + {15}^{0} ) - tan( \theta - {15}^{0}) }$

$\implies \tt \: 2= \frac{tan( \theta + {15}^{0} ) + tan( \theta - {15}^{0}) }{tan( \theta + {15}^{0} ) - tan( \theta - {15}^{0}) }$

We know that,

$\boxed{ \gray {\sf \: tan \: \theta = \frac{sin \: \theta}{cos \: \theta} } }$

$\implies \tt \: 2 = \frac{ \: \: \: \: \frac{sin(\theta + {15}^{0}) }{cos(\theta + {15}^{0} )} + \frac{sin(\theta - {15}^{0})}{cos(\theta - {15}^{0})} \: \: \: \: }{ \frac{sin(\theta + {15}^{0})}{cos(\theta + {15}^{0})} - \frac{sin(\theta - {15}^{0})}{cos(\theta - {15}^{0})} }$

$\tt \implies \: 2 = \frac{ \: \: \: \: \frac{sin(\theta + {15}^{0}).cos(\theta - {15}^{0}) + sin(\theta - {15}^{0}).cos(\theta + {15}^{0})}{cos(\theta + {15}^{0}).cos(\theta - {15}^{0})} \: \: \: \: }{ \frac{sin(\theta + {15}^{0}).cos(\theta - {15}^{0}) - sin(\theta - {15}^{0}).cos(\theta + {15}^{0})}{cos(\theta + {15}^{0}).cos(\theta - {15}^{0})} }$

$\tt \implies \: 2 = \frac{ \: \: \: \: \frac{sin(\theta + {15}^{0}).cos(\theta - {15}^{0}) + sin(\theta - {15}^{0}).cos(\theta + {15}^{0})}{ \cancel{cos(\theta + {15}^{0})}. \cancel{cos(\theta - {15}^{0})}} \: \: \: \: }{ \frac{sin(\theta + {15}^{0}).cos(\theta - {15}^{0}) - sin(\theta - {15}^{0}).cos(\theta + {15}^{0})}{ \cancel{cos(\theta + {15}^{0})}. \cancel{cos(\theta - {15}^{0})}} }$

$\tt \implies \: 2 = \frac{sin(\theta + {15}^{0}).cos(\theta - {15}^{0}) + sin(\theta - {15}^{0}).cos(\theta + {15}^{0})}{sin(\theta + {15}^{0}).cos(\theta - {15}^{0}) - sin(\theta - {15}^{0}).cos(\theta + {15}^{0})}$

We know that,

$\boxed{ \gray {\sf \: sin(A+B) = sinA \: cosB + cosA \: sinB} }\\ { \boxed{{\gray{ \sf{sin(A-B) = sinA \: cosB - cosA \: sinB}}}}}$

so,

$\sf \: sin(\theta + 15°) \: cos(\theta - 15°) = sinA \: cosB$

therefore,

$\tt \implies \: 2 = \frac{sin((\theta + {15}^{0} ) + (\theta - {15}^{0} ))}{sin((\theta + {15}^{0} ) - (\theta - {15}^{0}) }$

$\tt \implies \: 2 = \frac{sin(\theta + {15}^{0} + \theta - {15}^{0} )}{sin(\theta + {15}^{0} - \theta - {15}^{0}) }$

$\tt \implies \: 2 = \frac{sin(\theta + \cancel{ {15}^{0} }+ \theta - \cancel{{15}^{0} })}{sin( \cancel{\theta } + {15}^{0} - \cancel{\theta} - {15}^{0}) }$

$\tt \implies \: 2 = \frac{sin(2 \theta)}{sin \: {30}^{0} }$

$\tt \implies \: 2 = \frac{ \: sin \: 2\theta}{ \frac{1}{2} }$

$\tt \implies \: 2 = 2 \: sin \: 2 \theta$

$\tt \implies \: \frac{ \cancel{2}}{ \cancel{2}} = sin \: 2 \theta$

$\tt \implies \: 1 = sin \: 2 \theta$

We know that,

sin 90° = 1

$\tt \implies \: 2 \theta = {90}^{0}$

$\tt \implies \theta = \frac{ \: \: \: \: \: \: { \cancel{90} {}^{0} } \: ^{45} }{ \cancel2} \\ \\ \boxed{ \red{ \tt \implies \theta \: = {45}^{0} }}$