Prove the following identities:$ (1+\cot A-\operatorname{cosec} A)(1+\tan A+\sec A) = 2 $

To do:

We have to prove that \( (1+\cot A-\operatorname{cosec} A)(1+\tan A+\sec A) = 2 \).

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,

$(1+\cot A-\operatorname{cosec} A)(1+\tan A+\sec A)=\left(1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{1}{\cos A}\right)$

$=\left(\frac{\sin A+\cos A-1}{\sin A}\right)\left(\frac{\cos A+\sin A+1}{\cos A}\right)$

$=\frac{[(\sin A+\cos A)-1][(\sin A+\cos A)+1]}{\sin A \cos A}$

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

$=\frac{\sin ^{2} A+\cos ^{2} A+2 \sin A \cos A-1}{\sin A \sin A}$

$=\frac{1+2 \sin A \cos A-1}{\sin A \cos A}$

$=\frac{2 \sin A \cos A}{\sin A \cos A}$

$=2$

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Updated on: 10-Oct-2022

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