Question

1+ cosA by sinA+ sinA by 1+cosA=2cosceA​

Answer 1

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Answer 2

Step-by-step explanation:

$\frac{1 + \cos(A)}{ \sin(A) } + \frac{ \sin(A)}{1 + \cos(A) } \\ ⇒ \frac{(1 + {cos}A)^{2} + {sin}^{2}A}{sinA(1 +cos A)} \\ ⇒ \frac{ {1}^{2} + 2 \times 1 \times cos A + {cos}^{2}A + {sin }^{2}A}{sinA(1 + cosA) } \\⇒ \frac{1 + 2cos A + {cos}^{2} A + sin^{2} A }{sinA(1 + cos A )} \\ ⇒ \frac{1 + 2cosA + 1 }{sin A(1 + cosA) } \\ ⇒ \frac{2 + 2cos A }{sinA(1 + cos A )} \\ ⇒\frac{2(1 + cos A)}{sinA(1 + cosA) } \\ ⇒ \frac{2}{sin A } \\ ⇒ 2 \times \frac{1}{sinA} \\ ⇒2 \times cosecA \\ ⇒ 2cosecA \\ \\\red{\bold {\underline{\underline{Answer:}}}}2cosecA$

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