Question

3.
Prove that (cosec A-sin A) (sec A-cos A)=1/tanA+cotA
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Answer 1

Question:

$\sf{Prove\:that}\\\sf{\left(cosecA-sinA\right)\left(secA-cosA\right)=\dfrac{1}{tanA+cotA}}$

Identities Used:

$\sf{cosecA=\dfrac{1}{sinA}}\\\sf{secA=\dfrac{1}{cosA}}\\\sf{1-sin^2A=cos^2A}\\\sf{1-cos^2A=sin^2A}\\\sf{sin^2A+cos^2A=1}\\\sf{tanA=\dfrac{sinA}{cosA}}\\\sf{cotA=\dfrac{cosA}{sinA}}$

Solution:

$\sf{Taking\:L.H.S.}$

$\sf{=\left(cosecA-sinA\right)\left(secA-cosA\right)}$

$\sf{=\left(\dfrac{1}{sinA}-sinA\right)\left(\dfrac{1}{cosA}-cosA\right)}$

$\sf{=\left(\dfrac{1-sin^2A}{sinA}\right)\left(\dfrac{1-cos^2A}{cosA}\right)}$

$\sf{=\dfrac{cos^2A.sin^2A}{cosA.sinA}}$

$\sf{=sinA.cosA}$

$\sf{Taking\:R.H.S.}$

$\sf{\dfrac{1}{\dfrac{sinA}{cosA}+\dfrac{cosA}{sinA}}}$

$\sf{\dfrac{sinA.cosA}{sin^2A+cos^2A}}$

$\sf{sinA.cosA}$

$\mathrm{L.H.S.=R.H.S.}$

$\sf{\bf{HENCE\:PROVED}}$

Answer 2

Refer to the attachment above for your answer!!✔✔