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

To do:

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

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,

Let us consider LHS,

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

$=\frac{\cos \theta-\sin \theta}{\sin \theta \cos \theta} \times \frac{\cos ^{2} \theta-\sin ^{2} \theta}{\sin \theta \cos \theta}$

$=\frac{(\cos \theta-\sin \theta)\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta \cos ^{2} \theta}$

$=\frac{(\cos \theta-\sin \theta)[(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)]}{\sin ^{2} \theta \cos ^{2} \theta}$

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

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

Let us consider RHS,

$(\operatorname{cosec} \theta+\sec \theta)(\sec \theta \operatorname{cosec} \theta-2)=\left(\frac{1}{\sin \theta}+\frac{1}{\cos \theta}\right)\left(\frac{1}{\cos \theta \sin\theta}-2\right)$

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

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

Here,

LHS $=$ RHS

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Updated on: 10-Oct-2022

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