Math, asked by sohanlal210067, 1 year ago

cot theta upon 1 minus 10 theta + tan theta 1 minus cot theta is equal to 1 + cos theta sec theta ​

Answers

Answered by ShuchiRecites
12

Solution: cot∅/(1 - tan∅) + tan∅/(1 - cot∅)

→ cos∅/sin∅/(1 - sin∅/cos∅) + sin∅/cos∅/(1 - cos∅/sin∅)

→ cos∅/sin∅/(cos∅ - sin∅)/cos∅ + sin∅/cos∅/(sin∅ - cos∅)/sin∅

→ cos²∅/sin∅(cos∅ - sin∅) + sin²∅/cos∅(sin∅ - cos∅)

→ cos²∅/sin∅(cos∅ - sin∅) + sin²∅/[- cos∅(cos∅ - sin∅)]

→ 1/(cos∅ - sin∅) × [cos²∅/sin∅ - sin²∅/cos∅]

→ 1/(cos∅ - sin∅) × [(cos³∅ - sin³∅)/sin∅ cos∅]

→ 1/(cos∅ - sin∅) × (cos∅ - sin∅)(cos²∅ + cos∅ sin∅ + sin²∅)/cos∅ sin∅

→ (1 + cos∅ sin∅)/sin∅ cos∅

→ cosec∅ sec∅ + 1

Hence Proved

Answered by Sharad001
141

Question :-

Prove that ,

 \sf{ \footnotesize \frac{ \green{\cot( \theta)} }{\red{1 -  \tan( \theta) }}  +  \frac{ \tan( \theta) }{1 -  \cot( \theta) }  = 1 +\pink{  \csc( \theta)  \:  \sec( \theta) }}  \\ \\

Formula used :-

 \star \red{  \tan( \theta)  }=  \frac{ \green{ \sin( \theta) }}{ \cos( \theta) }  \\    \star  \pink{  \cot( \theta) } =  \frac{ \cos( \theta) }{  \red{\sin( \theta) }}  \\  \star \:   \green{{x}^{3}  -  {y}^{3} } = \red{ (x - y)( {x}^{2}  +  {y}^{2}  + xy)}

Proof :-

We have to prove LHS = RHS

Taking left hand side ,

  \rightarrow \footnotesize \frac{ \cot( \theta) }{1 -  \tan( \theta)}  + \frac{ \tan( \theta) }{1 -  \cot( \theta) }\\  \\  \rightarrow  \frac{ \frac{ \cos( \theta) }{ \sin( \theta) } }{1 -  \frac{ \sin( \theta) }{ \cos( \theta) } }  +  \frac{ \frac{ \sin( \theta) }{  \cos( \theta) } }{1- \frac{ \cos( \theta) }{ \sin( \theta) }}   \\  \\  \rightarrow \footnotesize  \frac{ { \cos }^{2}( \theta) }{ \sin( \theta) \big(  \cos( \theta)  -  \sin( \theta) \big) }  +  \frac{ { \sin }^{2}( \theta) }{ \cos( \theta) \big( \sin( \theta)  -  \cos( \theta)   \big)} \:   \\  \\  \rightarrow \footnotesize \: \frac{ { \cos }^{2}( \theta) }{ \sin( \theta) \big(  \cos( \theta)  -  \sin( \theta) \big) } \:  - \frac{ { \sin }^{2}( \theta) }{ \cos( \theta) \big(  - \sin( \theta)   +   \cos( \theta)   \big)} \:  \\  \\  \rightarrow \footnotesize  \frac{ { \cos}^{3} ( \theta) -  { \sin }^{3} ( \theta)}{ \big( \sin( \theta)  \cos( \theta) \big) \big( \cos( \theta)   -  \sin( \theta) \big)  }  \\  \\  \rightarrow \footnotesize \frac{ \big( \cos( \theta) -  \sin( \theta) \big) \big( { \sin }^{2}( \theta)   +  { \cos }^{2}( \theta ) +  \cos( \theta)  \sin( \theta)   \big)  }{ \big( \sin( \theta)  \cos( \theta) \big) \big( \cos( \theta)   -  \sin( \theta) \big) \: } \\  \\  \rightarrow \:  \frac{1 +  \cos( \theta)  \sin( \theta) }{ \cos( \theta) \sin( \theta)  }  \\  \\  \rightarrow \:  \frac{1}{ \cos( \theta) }  \frac{}{ \sin( \theta) }  + 1 \\  \\  \rightarrow \: 1 +  \sec( \theta)  \csc( \theta)

Hence proved .

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