Question

cotA-cosA/cotA+cosA=cosecA-1/cosecA+1​

Answer 1

$\large{\underline{\underline{\red{\bf{Solution:}}}}}\\ \\ \\ {\underline{\bf Given:}}\\ \\ \\ \longrightarrow \sf \dfrac{\cot A-\cos A}{\cot A+\cos A}=\dfrac{cosec\;A-1}{cosec\;A+1}\\ \\ \\ {\underline{\bf Now,\;we\;will\;take\;LHS\;part,}}\\ \\ \\ \longrightarrow \sf \dfrac{\cot A-\cos A}{\cot A+\cos A}\\ \\ \\ \bf Now,\;we\;know\;that,\;\cot A = \dfrac{\cos A}{\sin A}.\\ \\ \\ \longrightarrow \sf \dfrac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{\cos A}{\sin A}+\cos A}$

${\underline{\bf Now,\;take\;LCM,}}\\ \\ \\ \longrightarrow \sf \dfrac{\cos A-\cos A\;\sin A}{\cos A+\cos A\;\sin A}\\ \\ \\ {\underline{\bf Now,\;divide\;both\;numerator\;and\;denominator\;by\;\sin A\;\cos A,\;we\;get}}\\ \\ \\ \longrightarrow \sf \dfrac{\dfrac{\cos A-\cos A\;\sin A}{\sin A\;\cos A}}{\dfrac{\cos A+\cos A\;\sin A}{\sin A\;\cos A}}\\ \\ \\ \longrightarrow \sf \dfrac{\cos A-1}{\cos A+1}\\ \\ \\ \bf We\;know\;that\;\cos A=\dfrac{1}{\sin A}$

$\longrightarrow \sf \dfrac{\dfrac{1}{\sin A}-1}{\dfrac{1}{\sin A}+1}\\ \\ \\ \longrightarrow \sf \dfrac{cosec\;-1}{cosec\;+1} \\ \\ \\ {\underline{\underline{\bf Hence\;Proved!!!}}}$

Answer 2

LHS=

Express in terms of sinA and cos A

Then divide the whole thing by cosA

$\frac{cotA-cosA}{cotA+cosA\\}\\ \\\\\frac{cosA/sinA -cosA}{cosA/sinA +cosA}\\ \\\frac{1/sinA -1}{1/sinA +1}\\ \\ \frac{cosecA-1}{cosecA+1}$then you know 1/sinA = cosecA

and then u get the answer

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