Question

tan A/1- cot A+cotA/1-tanA
Prove that
= 1+tan A+ cot A.

​

Answer 1

$\boxed{ \purple{ \bf \:Prove that \: : \dfrac{tanA}{1 - cotA} + \dfrac{cotA}{1 - tanA} = \:1 + tanA + cotA }}$

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\tt \:  ⟼ cotA = \dfrac{1}{tanA}$

$\tt \:  \longrightarrow \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )$

$\large\underline\purple{\bold{Solution :- }}$

☆ Consider LHS,

$\tt \:  \longrightarrow \: \dfrac{tanA}{1 - cotA} + \dfrac{cotA}{1 - tanA}$

$\tt \:  \longrightarrow \: \dfrac{tanA}{1 - \dfrac{1}{tanA} } - \dfrac{1}{tanA(1 - tanA)}$

$\tt \:  \longrightarrow \: \dfrac{ {tan}^{2} A}{tanA - 1} + \dfrac{1}{tanA(1 - tanA)}$

$\tt \:  \longrightarrow \: \dfrac{ {tan}^{2} A}{tanA - 1} - \dfrac{1}{tanA(tanA - 1)}$

$\tt \:  \longrightarrow \: \dfrac{ {tan}^{3} A - 1}{tanA(tanA - 1)}$

$\tt \:  \longrightarrow \: \dfrac{ \cancel{(tanA - 1)} \: \: ( {tan}^{2}A + tanA + 1) }{tanA \: \: \cancel{(tanA - 1)}}$

$\tt \:  \longrightarrow \: \dfrac{ {tan}^{2}A + tanA + 1 }{tanA}$

$\tt \:  \longrightarrow \: \dfrac{ {tan}^{2} A}{tanA} + \dfrac{tanA}{tanA} + \dfrac{1}{tanA}$

$\tt \:  \longrightarrow \: tanA + 1 + cotA$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\Large{\bold{\purple{\underline{More \: InFoRmAtIoN\::}}}} \\ \end{gathered}$

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

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