Question

My last questionJohncarl help​

Answer 1

LHS

$\\ \sf\longmapsto \dfrac{1}{sec\Theta+\tan\Theta}$

• sec^2A-tan^2A=1

$\\ \sf\longmapsto \dfrac{sec^2\Theta-tan^2\Theta}{sec\Theta+tan\:Theta}$

• a^2-b^2=(a+b)(a-b)

$\\ \sf\longmapsto \dfrac{(sec\Theta+tan\Theta)(sec\Theta-tan\Theta)}{(sec\Theta+tan\:Theta)}$

$\\ \sf\longmapsto sec\Theta-tan\Theta$

$\\ \sf\longmapsto \dfrac{1}{cos\Theta}-\dfrac{sin\Theta}{cos\Theta}$

$\\ \sf\longmapsto \dfrac{1-sin\Theta}{cos\:Theta}$

Answer 2

Answer:

• Hope its help❤✨✨

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