Prove the following identities, where the angles involved are acute an

1.	Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
Answer» Step-by-step EXPLANATION: $\underline{\underline{\maltese\: \: \textbf{\textsf{To\ Prove}}}}$ ${\boxed{\sf{(cosec\ \theta\ -\ cot\ \theta)^2\ =\ \dfrac{1\ -\ cos\ \theta}{1\ +\ cos\ \theta}}}}$ $\underline{\underline{\maltese\: \: \textbf{\textsf{Solution}}}}$ ➤ By SOLVING the L.H.S ➤ That is, $\sf : \implies {(cosec\ \theta\ -\ cot\ \theta)^2}$ We need to MAKE it in TERMS of cos θ and SIN θ. $\sf : \implies {\bigg(\dfrac{1}{sin\ \theta}\ -\ \dfrac{cos\ \theta}{sin\ \theta} \bigg)^2}$ $\\$ $\sf : \implies {\bigg(\dfrac{1\ -\ cos\ \theta}{sin\ \theta} \bigg)^2}$ $\\$ $\sf : \implies {\dfrac{(1\ -\ cos\ \theta)^2}{sin^2\ \theta}}$ $\\$ $\begin{gathered} \small\boxed{ \begin{array}{cc} \sf{We\ know\ that,} \\ \\ \sf {cos^2\ \theta\ +\ sin^2\ \theta\ =\ 1} \\ \\ \sf {sin^2\ \theta\ =\ 1\ -\ cos^2\ \theta} \end{array} } \end{gathered}$ $\\$ $\sf : \implies {\dfrac{(1\ -\ cos\ \theta)^2}{1\ -\ cos^2\ \theta}}$ $\\$ $\sf : \implies {\dfrac{(1\ -\ cos\ \theta)^2}{(1\ -\ cos\ \theta)^2}}$ $\\$ $\begin{gathered} \small\boxed{ \begin{array}{cc} \sf {Using,\ a^2\ -\ b^2\ =\ (a\ -\ b)\ (a\ +\ b)} \\ \\ \sf {Putting,\ a\ =\ 1,\ b\ =\ cos\ \theta} \end{array} } \end{gathered}$ $\\$ $\sf : \implies {\dfrac{(1\ -\ cos\ \theta)^2}{(1\ +\ cos\ \theta)(1\ -\ cos\ \theta)}}$ $\\$ $\sf : \implies {\dfrac{\cancel {(1\ -\ cos\ \theta)}(1\ -\ cos\ \theta)}{(1\ +\ cos\ \theta) \cancel {(1\ -\ cos\ \theta)}}}$ $\\$ $\sf : \implies {\dfrac{1\ -\ cos\ \theta}{1\ +\ cos\ \theta}}$ $\\$ $\sf : \implies {R.H.S}$ $\\$ $\large {\therefore{\boxed{\boxed{\sf {(cosec\ \theta\ -\ cot\ \theta)^2\ =\ \dfrac{1\ -\ cos\ \theta}{1\ +\ cos\ \theta}}}}}}$ $\\$ $\underline{\textbf{\textsf{Hence\ Proved...!!!}}}$

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer»

Step-by-step EXPLANATION:

$\underline{\underline{\maltese\: \: \textbf{\textsf{To\ Prove}}}}$

${\boxed{\sf{(cosec\ \theta\ -\ cot\ \theta)^2\ =\ \dfrac{1\ -\ cos\ \theta}{1\ +\ cos\ \theta}}}}$

$\underline{\underline{\maltese\: \: \textbf{\textsf{Solution}}}}$

➤ By SOLVING the L.H.S

➤ That is,

$\sf : \implies {(cosec\ \theta\ -\ cot\ \theta)^2}$

We need to MAKE it in TERMS of cos θ and SIN θ.

$\sf : \implies {\bigg(\dfrac{1}{sin\ \theta}\ -\ \dfrac{cos\ \theta}{sin\ \theta} \bigg)^2}$

$\\$

$\sf : \implies {\bigg(\dfrac{1\ -\ cos\ \theta}{sin\ \theta} \bigg)^2}$

$\\$

$\sf : \implies {\dfrac{(1\ -\ cos\ \theta)^2}{sin^2\ \theta}}$

$\\$

$\begin{gathered} \small\boxed{ \begin{array}{cc} \sf{We\ know\ that,} \\ \\ \sf {cos^2\ \theta\ +\ sin^2\ \theta\ =\ 1} \\ \\ \sf {sin^2\ \theta\ =\ 1\ -\ cos^2\ \theta} \end{array} } \end{gathered}$

$\\$

$\sf : \implies {\dfrac{(1\ -\ cos\ \theta)^2}{1\ -\ cos^2\ \theta}}$

$\\$

$\sf : \implies {\dfrac{(1\ -\ cos\ \theta)^2}{(1\ -\ cos\ \theta)^2}}$

$\\$

$\begin{gathered} \small\boxed{ \begin{array}{cc} \sf {Using,\ a^2\ -\ b^2\ =\ (a\ -\ b)\ (a\ +\ b)} \\ \\ \sf {Putting,\ a\ =\ 1,\ b\ =\ cos\ \theta} \end{array} } \end{gathered}$