Question

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

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

Answer 1

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.

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

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

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

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

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

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

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

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

$\sf : \implies {\dfrac{\cancel {(1\ -\ cos\ \theta)}(1\ -\ cos\ \theta)}{(1\ +\ cos\ \theta) \cancel {(1\ -\ cos\ \theta)}}}$

$\sf : \implies {\dfrac{1\ -\ cos\ \theta}{1\ +\ cos\ \theta}}$

$\sf : \implies {R.H.S}$

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

$\underline{\textbf{\textsf{Hence\ Proved...!!!}}}$