Math, asked by shaukataziz261, 6 months ago

prove that : 1/tanA - 1/tan2A = cosec2A ​

Answers

Answered by anshulgoel16
4

Answer:

1/sinA/cosA - 1/sin2A/cos2A

cosA/sinA - cos2A/sin2A

cotA-cot2A=cosec2A

Answered by Mysterioushine
46

 \huge {\mathfrak{ \underline{  \red s \green o  \blue l \pink u \purple t \orange  i o \green n  }}} : -

 \dfrac{1}{ \tan(A) }  -  \dfrac{1}{ \tan(2A) }

 \large \underline {\bold {\boxed{ \tan(2A)  =  \dfrac{2 \tan(A) }{1 +  \tan {}^{2} (A) } }}}

  \longrightarrow \dfrac{1}{ \tan(A) }  -  \dfrac{1}{ \dfrac{2 \tan(A) }{1 +   \tan {}^{2} (A)  } }  \\  \\   \longrightarrow  \frac{1}{ \tan(A) }  -  \dfrac{1 +  \tan {}^{2} (A) }{2 \tan(A) } \\  \\  \longrightarrow \:  \frac{2 - 1 +  \tan {}^{2} (A) }{2 \tan(A) }  \\  \\  \longrightarrow \:  \frac{1 +  \tan {}^{2} (A) }{2 \tan(A) }

 \large {\underline {\bold {\boxed{ \sin(2A)  =  \frac{2 \tan(A) }{1 + \tan {}^{2} (A)  } }}}}

 \longrightarrow \:  \dfrac{1}{ \sin(2A) }

 \large {\underline {\bold {\boxed{ \cosec(A)  =  \dfrac{1}{ \sin(A) } }}}}

 \longrightarrow \:  \cosec(2A)

 \dag \:  \huge \bold{ \green{\boxed{  \rm{ \blue{ Hence \: proved}}}}}

 \huge \underline \mathfrak{  \red s \blue o \green m \pink e \:  \:  \red f \green  o \blue r \pink m \purple u  \orange la \red e } : -

(1) \sf  \:   \: \sin {}^{2} \theta +  \cos {}^{2}  \theta  = 1 \\  \\ (2) \:  \cosec {}^{2}  \theta  -  \cot {}^{2} \theta = 1 \\  \\ (3) \:  \sec {}^{2} \theta  -  \tan {}^{2} \theta = 1 \\  \\ (4) \cos2 \theta = 2 \cos {}^{2} \theta - 1 \\  \\ (5) \sin 2\theta = 2 \sin\theta\cos\theta

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