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Prove the following trigonometric identities:$ (\operatorname{cosec} A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1 $
To do:
We have to prove that \( (\operatorname{cosec} A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1 \).
Solution:
We know that,
$\tan A=\frac{\sin A}{\cos A}$.....(i)
$\cot A=\frac{\cos A}{\sin A}$.....(ii)
$\operatorname{cosec} A=\frac{1}{\sin A}$.....(iii)
$\sec A=\frac{1}{\cos A}$.....(iv)
$\sin^2 A+\cos ^{2} A=1$.......(v)
Therefore,
$(\operatorname{cosec} A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A})$
$=(\frac{1-\sin^2 A}{\sin A})(\frac{1-cos^2A}{\cos A})(\frac{\sin^2 A+\cos^2A}{\sin A\cos A})$
$=(\frac{\cos^2A}{\sin A})(\frac{sin^2A}{\cos A})(\frac{1}{\sin A\cos A})$
$=\frac{\sin A\cos A}{\sin A\cos A}$
$=1$
Hence proved.
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