Prove: $(cosecA-sinA)(secA-cosA)=\frac{1}{tanA+cotA}$.

Given: $(cosecA-sinA)(secA-cosA)=\frac{1}{tanA+cotA}$.

To do: To prove: $(cosecA-sinA)(secA-cosA)=\frac{1}{tanA+cotA}$.

Solution:

$(cosecA-sinA)(secA-cosA)=\frac{1}{tanA+cotA}$

$L.H.S.=(cosecA-sinA)(secA-cosA)$

$=( \frac{1}{sinA}-sinA)-( \frac{1}{cosA}-cosA)$           [$\because cosecA=\frac{1}{sinA}$ and $secA=\frac{1}{cosA}$]

$=( \frac{1-sin^2A}{sinA})( \frac{1-cos^2A}{cosA})$

$=( \frac{cos^2A}{sinA})( \frac{sin^2A}{cosA})$          [$\because 1-sin^2A=cos^2A$ and $1-cos^2A=sin^2A$]

$=sinAcosA$

And $R.H.S.=\frac{1}{tanA+cotA}$

$=\frac{1}{\frac{sinA}{cosA}+\frac{cosA}{sinA}}$   [$\because tanA=\frac{sinA}{cosA}$ and $cotA=\frac{cosA}{sinA}$]

$=\frac{1}{\frac{sin^2A+cos^2}{cosAsinA}}$      [$\because sin^2A+cos^2A=1$]

$=sinAcosA$

Thus, $L.H.S.=R.H.S.$

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Updated on: 10-Oct-2022

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