Question

(cotA+cosecA)^2= (1+cosA)/(1-cosA) prove it.​

Answer 1

Explanation :

Refer to the attachment.

Formulas related trignometry :

★ CosA/SinA = CotA

★ SinA/CosA = TanA

★ 1/CosecA = SinA

★ 1/SinA = CosecA

★ 1/CosA = SecA

★ 1/SecA = CosA

★ Sin²A + Cos²A = 1

★ ( a² - b² ) = (a+b)(a-b)

Answer 2

$\bf To\ prove:\ (cot\ A\ +\ cosec\ A)^2\ =\ \dfrac{(1\ +\ cos\ A)}{(1\ -\ cos\ A)}\\\\\\Proof:\\\\\\\sf Left\ Hand\ Side:\\\\\\(cot\ A\ +\ cosec\ A)^2\\\\\\\tt We\ know\ that\ cot\ \theta\ =\ \dfrac{cos\ \theta}{sin\ \theta}\ and\ cosec\ \theta\ =\ \dfrac{1}{sin\ \theta}.\\\\\\\implies\ \sf\ \bigg(\dfrac{cos\ A}{sin\ A}\ +\ \dfrac{1}{sin\ A}\bigg)^2\\\\\\\\=\ \ \ \ \ \bigg(\dfrac{cos\ A\ +\ 1}{sin\ A}\bigg)^2\\\\\\\\=\ \ \ \ \ \bigg(\dfrac{(cos\ A\ +\ 1)^2}{sin^2\ A}\bigg)$

$\sf =\ \ \ \ \ \dfrac{(cos\ A\ +\ 1)(cos\ A\ +\ 1)}{sin^2\ A}\\\\\\\tt We\ know\ that\ sin^2\ \theta\ +\ cos^2\ \theta\ =\ 1.\\\\Hence, sin^2\ \theta\ =\ 1\ -\ cos^2\ \theta.\\\\\\\sf =\ \ \ \ \ \ \dfrac{(cos\ A\ +\ 1)(cos\ A\ +\ 1)}{1\ -\ cos^2\ A}\\\\\\\\=\ \ \ \ \ \ \dfrac{(cos\ A\ +\ 1)(cos\ A\ +\ 1)}{(1\ -\ cos\ A)(1\ +\ cos\ A)}\ \ \ \ \ \ \ \ \ \ \ \ \ \bigg[a^2\ -\ b^2\ =\ (a\ +\ b)(a\ -\ b)\bigg]$

$\sf =\ \ \ \ \ \dfrac{1\ +\ cos\ A}{1\ -\ cos\ A}\\\\\\\\\bf Hence,\ proved.$