Question

My last question
Johncarl help​

Answer 1

LHS

`\\ \sf\longmapsto (1)/(sec\Theta+\tan\Theta)`

• sec^2A-tan^2A=1

`\\ \sf\longmapsto (sec^2\Theta-tan^2\Theta)/(sec\Theta+tan\:Theta)`

• a^2-b^2=(a+b)(a-b)

`\\ \sf\longmapsto ((sec\Theta+tan\Theta)(sec\Theta-tan\Theta))/((sec\Theta+tan\:Theta))`

`\\ \sf\longmapsto sec\Theta-tan\Theta`

`\\ \sf\longmapsto (1)/(cos\Theta)-(sin\Theta)/(cos\Theta)`

`\\ \sf\longmapsto (1-sin\Theta)/(cos\:Theta)`

Answer 2

• Hope its help❤✨✨

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