Question

trignometry solve this pls it is compound related ques.



prove that​

Answer 1

$\huge \red{\mathfrak{solution}}$

To prove :

$tan(45 - \alpha ) = \frac{cos (\alpha ) - \sin( \alpha ) }{ \cos( \alpha ) + \sin( \alpha ) }$

we have to use formula :

tan(a-b) =( tana - tanb/ 1+ tana tanb)

LHS

$\rightarrow\tan(45 - \alpha ) \\ \\ \rightarrow \: \frac{ \tan(45) - \tan( \alpha ) }{1 + \tan(45) \tan( \alpha ) } \\ \\ tan45 = 1$

$\rightarrow \: \frac{1 - \tan( \alpha ) }{1 + \tan( \alpha ) } \\ \\ \rightarrow \: \frac{1 - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } }{1 + \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } \\ \\ \rightarrow \: \frac{ \frac{ \cos( \alpha ) - \sin( \alpha ) }{ \cos( \alpha ) } }{ \frac{ \cos( \alpha ) + \sin( \alpha ) }{ \cos( \alpha ) } }$

Here cos@ gets cancel out

$\rightarrow \: \huge{ \frac{ \cos( \alpha ) - \sin( \alpha ) }{ \cos( \alpha ) + \sin( \alpha )} }$

LHS= RHS

$\green{ \huge \: \mathfrak{proved}}$

Answer 2

AnswEr :

• To Prove :

$\sf \tan(45 - \theta) = \dfrac{ \cos( \theta) - \sin( \theta) }{\cos( \theta) + \sin( \theta)}$

• Proof :

$\longrightarrow \large \sf \tan(45 - \theta)$

⠀⠀⠀⠀⋆ $\small \sf \tan(a - b) = \frac{ \tan(a) - \tan(b) }{1 + \tan(a) \tan(b) }$

$\longrightarrow \large\sf \dfrac{ \tan(45) - \tan( \theta) }{1 + \tan(45) \tan( \theta) }$

⠀⠀⠀⠀⋆ $\small \sf \tan(45)=1$

$\longrightarrow \large\sf \dfrac{ 1 - \tan( \theta) }{1 + (1 \times \tan( \theta)) }$

$\longrightarrow \large\sf \dfrac{ 1 - \tan( \theta) }{1 + \tan( \theta) }$

$\longrightarrow \large\sf \dfrac{ 1 - \frac{ \sin(\theta) }{ \cos(\theta) } }{1 + \frac{ \sin(\theta) }{ \cos(\theta) } }$

$\longrightarrow \large\sf \dfrac{\frac{ \cos(\theta) - \sin(\theta) }{ \cancel{\cos(\theta)} } }{\frac{ \cos(\theta) + \sin(\theta) }{ \cancel{\cos(\theta)} } }$

$\longrightarrow \large \boxed{ \sf \dfrac{\cos(\theta) - \sin(\theta)}{\cos(\theta) + \sin(\theta)} }$