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JohnCarI

Answer:

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LHS

[tex]\\ \sf\longmapsto \dfrac{1}{sec\Theta+\tan\Theta}[/tex]

  • sec^2A-tan^2A=1

[tex]\\ \sf\longmapsto \dfrac{sec^2\Theta-tan^2\Theta}{sec\Theta+tan\:Theta}[/tex]

  • a^2-b^2=(a+b)(a-b)

[tex]\\ \sf\longmapsto \dfrac{(sec\Theta+tan\Theta)(sec\Theta-tan\Theta)}{(sec\Theta+tan\:Theta)}[/tex]

[tex]\\ \sf\longmapsto sec\Theta-tan\Theta[/tex]

[tex]\\ \sf\longmapsto \dfrac{1}{cos\Theta}-\dfrac{sin\Theta}{cos\Theta}[/tex]

[tex]\\ \sf\longmapsto \dfrac{1-sin\Theta}{cos\:Theta}[/tex]

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