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Prove the following identities:$ (\sec A-\operatorname{cosec} A)(1+\tan A+\cot A)=\tan A \sec A-\cot A \operatorname{cosec} A $
To do:
We have to prove that \( (\sec A-\operatorname{cosec} A)(1+\tan A+\cot A)=\tan A \sec A-\cot A \operatorname{cosec} A \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
Let us consider LHS,
$(\sec A-\operatorname{cosec} A)(1+\tan A+\cot A)=\left(\frac{1}{\cos A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)$
$=\frac{\sin A-\cos A}{\sin A \cos A} \times \frac{\sin A \cos A+\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}$
$=\frac{\sin A-\cos A}{\sin A \cos A} \times \frac{1+\sin A \cos A}{\sin A \cos A}$
$=\frac{(\sin A-\cos A)(1+\sin A \cos A)}{\sin ^{2} A \cos ^{2} A}$
Let us consider RHS,
$\tan A \sec A-\cot A \operatorname{cosec} A=\frac{\sin A}{\cos A} \times \frac{1}{\cos A}-\frac{\cos A}{\sin A} \times \frac{1}{\sin A}$
$=\frac{\sin A}{\cos ^{2} A}-\frac{\cos A}{\sin ^{2} A}$
$=\frac{\sin ^{3} A-\cos ^{3} A}{\sin ^{2} A \cos ^{2} A}$
$=\frac{(\sin A-\cos . A)\left(\sin ^{2} A+\cos ^{2} A+\sin A \cos A\right)}{\sin ^{2} A \cos ^{2} A}$
$=\frac{(\sin A-\cos A)(1+\sin A \cos A)}{\sin ^{2} A \cos ^{2} A} $
Here,
LHS $=$ RHS
Hence proved.
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