Math, asked by paramesh24, 11 months ago

prove tan + 1/tan = sec.cosec​

Answers

Answered by sprao53413
2

Answer:

Please see the attachment

Attachments:
Answered by kaushik05
20

 \huge \red{ \mathfrak{solution}}

 \tan( \alpha )  +  \frac{1}{ \tan( \alpha ) }  =  \sec( \alpha )  \csc( \alpha )

LHS

 \tan( \alpha )  +  \frac{1}{ \tan( \alpha ) }  \\

 \rightarrow \:  \frac{ \sin( \alpha ) }{ \cos( \alpha ) }  +  \frac{1}{  \frac{ \sin( \alpha ) }{ \cos(  \alpha  ) }  }  \\  \\  \rightarrow \:   \frac{ \sin( \alpha ) }{ \cos( \alpha ) }  +  \frac{ \cos( \alpha ) }{ \sin( \alpha ) }  \\  \\  \rightarrow \:  \frac{ { \sin }^{2} (\alpha  )+  \cos ^{2} ( \alpha )  }{ \cos( \alpha ) \sin( \alpha )  }  \\  \\  \rightarrow \:  \frac{1}{  \cos( \alpha )  \sin( \alpha ) }  \\  \\  \rightarrow \:  \sec( \alpha )  \csc( \alpha )

LHS= RHS

 \huge \green{ \boxed{ \mathfrak{poved}}}

Formula used :

tan @= sin@/cos@

sin^2@+cos^2@ =1

cos@=1/sec@

sin@=1/ csc@

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