Question

(1+cosA/sinA)^2 = 1+cosA/1-sinA​

Answer 1

Answer:

Taking LHS:

$\sf \bf :\longmapsto \left(\dfrac{1+cosA}{sinA}\right)^2$

$\sf \bf :\longmapsto \dfrac{(1+cosA)^2}{sin^2A}$

$\sf \bf :\longmapsto \dfrac{(1+cosA)^2}{1-cos^2A}$

$\sf \bf :\longmapsto \dfrac{(1+cosA)(1+cosA)}{(1+cosA)(1-cosA}$

$\sf \bf :\longmapsto \dfrac{1+cosA}{1-cosA}$

= RHS

Identities used :

$\sf \bf sin^2A + cos^A = 1$

$a^2 - b^2 = (a + b)(a - b)$

Additional information:

$\sf \color{aqua}{Trigonometry\: Table}\\ \purple{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf \red{\angle A} & \red{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \red{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &\red{ \sf{90}^{ \circ}} \\ \hline \\ \rm \red{sin A} & \green{0} & \green{\dfrac{1}{2}}& \green{\dfrac{1}{ \sqrt{2} }} &\green{ \dfrac{ \sqrt{3}}{2} }&\green{1} \\ \hline \\ \rm \red{cos \: A} & \green{1} &\green{ \dfrac{ \sqrt{3} }{2}}&\green{ \dfrac{1}{ \sqrt{2} }} & \green{\dfrac{1}{2}} &\green{0} \\ \hline \\\rm \red{tan A}& \green{0} &\green{ \dfrac{1}{ \sqrt{3} }}&\green{1} & \green{\sqrt{3}} & \rm \green{\infty} \\ \hline \\ \rm \red{cosec A }& \rm \green{\infty} & \green{2}& \green{\sqrt{2} }&\green{ \dfrac{2}{ \sqrt{3} }}&\green{1} \\ \hline\\ \rm \red{sec A} & \green{1 }&\green{ \dfrac{2}{ \sqrt{3} }}& \green{\sqrt{2}} & \green{2} & \rm \green{\infty} \\ \hline \\ \rm \red{cot A }& \rm \green{\infty} & \green{\sqrt{3}}& \green{1} & \green{\dfrac{1}{ \sqrt{3} }} & \green{0}\end{array}}}}$