Question

Prove that $\dfrac{{\cos ec\left( {{{90}^ \circ } - x} \right)\sin \left( {{{180}^ \circ } - x} \right)\cot \left( {{{360}^ \circ } - x} \right)}}{{\sec \left( {{{180}^ \circ } + x} \right)\tan \left( {{{90}^ \circ } + x} \right)\sin \left( { - x} \right)}} = 1$

Answer 1

Hint: It is known that $\cos ec\left( {{{90}^ \circ } - x} \right) = \sec x$,$\sin \left( {{{180}^ \circ } - x} \right) = \sin x$,$\cot \left( {{{360}^ \circ } - x} \right) = - \cot x$, $\sec \left( {{{180}^ \circ } + x} \right) = - \sec x$,$\tan \left( {{{90}^ \circ } + x} \right) = - \cot x$and $\sin \left( { - x} \right) = - \sin x$. We do this conversion and simplify this if L.H.S will be equal to 1 then it will be proved that $L.H.S = R.H.S$.

Complete step-by-step answer:
\[L.H.S = \dfrac{{\cos ec\left( {{{90}^ \circ } - x} \right)\sin \left( {{{180}^ \circ } - x} \right)\cot \left( {{{360}^ \circ } - x} \right)}}{{\sec \left( {{{180}^ \circ } + x} \right)\tan \left( {{{90}^ \circ } + x} \right)\sin \left( { - x} \right)}}\]
Substitute $\cos ec\left( {{{90}^ \circ } - x} \right) = \sec x$,$\sin \left( {{{180}^ \circ } - x} \right) = \sin x$,$\cot \left( {{{360}^ \circ } - x} \right) = - \cot x$, $\sec \left( {{{180}^ \circ } + x} \right) = - \sec x$,$\tan \left( {{{90}^ \circ } + x} \right) = - \cot x$and $\sin \left( { - x} \right) = - \sin x$, we get,
$ \Rightarrow \dfrac{{\sec x\sin x\left( { - \cot x} \right)}}{{\left( { - \sec x} \right)\left( { - \cot x} \right)\left( { - \sin x} \right)}}$
As we can see every term get cancelled
$\therefore L.H.S = 1$
Given $R.H.S = 1$
Therefore, $L.H.S = R.H.S$proved

Note: Some important formula. Try to remember these formulas to solve these types of problems.

\[
  \sin ({90^ \circ } - x) = \cos x \\
  \tan ({90^ \circ } - x) = \cot x \\
  \cos ec({90^ \circ } - x) = \sec x \\
  \cos \left( {{{90}^ \circ } - x} \right) = \sin x \\
  \cot \left( {{{90}^ \circ } - x} \right) = \tan x \\
  \sec \left( {{{90}^ \circ } - x} \right) = \cos ecx \\
 \]
  \[
  \sin ({90^ \circ } + x) = \cos x \\
  \tan ({90^ \circ } + x) = - \cot x \\
  \cos ec({90^ \circ } + x) = \sec x \\
  \cos \left( {{{90}^ \circ } + x} \right) = - \sin x \\
  \cot \left( {{{90}^ \circ } + x} \right) = - \tan x \\
  \sec \left( {{{90}^ \circ } + x} \right) = - \cos ecx \\
 \]
$
  \sin ({180^ \circ } + x) = - \cos x \\
  \tan ({180^ \circ } + x) = \cot x \\
  \cos ec({180^ \circ } + x) = - \sec x \\
  \cos \left( {{{180}^ \circ } + x} \right) = - \sin x \\
  \cot \left( {{{180}^ \circ } + x} \right) = \tan x \\
  \sec \left( {{{180}^ \circ } + x} \right) = - \cos ecx \\
 $
 \[
  \sin ({360^ \circ } - x) = - \cos x \\
  \tan ({360^ \circ } - x) = - \cot x \\
  \cos ec({360^ \circ } - x) = - \sec x \\
  \cos \left( {{{360}^ \circ } - x} \right) = \sin x \\
  \cot \left( {{{360}^ \circ } - x} \right) = - \tan x \\
  \sec \left( {{{360}^ \circ } - x} \right) = \cos ecx \\
 \]