Prove that $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$
Prove that $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$
$L.H.S.$ $=\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x$
$=\cot x \cot 2 x-\cot 3 x(\cot 2 x+\cot x)$
$=\cot x \cot 2 x-\cot (2 x+x)(\cot 2 x+\cot x)$
$=\cot x \cot 2 x-\left[\frac{\cot 2 x \cot x-1}{\cot x+\cot 2 x}\right](\cot 2 x+\cot x)$
$\left[\because \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}\right]$
$=\cot x \cot 2 x-(\cot 2 x \cot x-1)=1=R .H .S.$
Similar Questions
If $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
and $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ then
If $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
and $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ then
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$\frac{{\cos A}}{{1 - \sin A}} = $
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
- [IIT 1988]
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$\sin {20^o}\,\sin {40^o}\,\sin {60^o}\,\sin {80^o} = $