Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Consider LHS

[tex]\rm \: \dfrac{ \cot \theta - \cos \theta}{ \cot \theta \: + \cos \theta}[/tex]

We know,

[tex]\boxed{ \rm{ \:cotx = \frac{cosx}{sinx} \: }} \\ [/tex]

So, using this, the above expression can be rewritten as

[tex]\rm \: = \: \dfrac{\dfrac{cos\theta }{sin\theta } - cos\theta }{\dfrac{cos\theta }{sin\theta } + cos\theta } \\ [/tex]

[tex]\rm \: = \: \dfrac{cos\theta \bigg(\dfrac{1}{sin\theta } - 1\bigg) }{cos\theta \bigg(\dfrac{1}{sin\theta } + 1\bigg)} \\ [/tex]

[tex]\rm \: = \: \dfrac{\dfrac{1}{sin\theta } - 1 }{\dfrac{1}{sin\theta } + 1} \\ [/tex]

[tex]\rm \: = \: \dfrac{cosec\theta - 1}{cosec\theta + 1} \\ [/tex]

Hence,

[tex]\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{ \cot \theta - \cos \theta}{ \cot \theta \: + \cos \theta} = \: \dfrac{cosec\theta - 1}{cosec\theta + 1} \: \: }} \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ [/tex]

Answer:

HERE IS YOUR ANSWER

Have a nice day/noon/night