Math, asked by dhaval1556, 1 year ago

tan2theta + cot2theta =??​

Answers

Answered by sanketj
1

 \:  \:  \:  \:  \: tan2x + cot2x \\  =tan2x +  \frac{1}{tan2x}   \:  \:  \:  \:  ...(cotx =  \frac{1}{tanx} ) \\  =  \frac{ {tan}^{2}2x  + 1}{tan2x}  \\  =  \frac{ {sec}^{2} 2x}{tan2x}  \:  \:  \:  \:  \:  \:  \:  \:  \:   \: ...(1 +  {tan}^{2} x =  {sec}^{2} x) \\  =  \frac{ \frac{1}{ {cos}^{2}2x } }{ \frac{sin2x}{cos2x} }   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \: ...(secx =  \frac{1}{cosx} \\  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  tanx =  \frac{sinx}{cosx} )  \\  = ( \frac{1}{ {cos}^{2} 2x} )( \frac{cos2x}{sin2x} )  =  \frac{1}{cos2x.sin2x}  \\  = sec2x.cosec2x \: \:  \:  \:  \:  \:  \:  ...(secx =  \frac{1}{cosx }  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   cosecx =  \frac{1}{sinx} )

Hence,

tan2A + cot2A = sec2A.cosec2A.

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