Question

bracket cosec - sin bracket sec - cos is equal to one upon tan A + cot a prove this identity

Answer 1

We have to prove that:

$(\csc A - \sin A)(\sec A - \cos A)=\frac{1}{\tan A+\cot A}$

Consider LHS:

$(\csc A - \sin A)(\sec A - \cos A)$

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

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

By using the trigonometric identity $\sin^2A+\cos^2A = 1$

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

= $\cos A \sin A$

Consider RHS:

$\frac{1}{\tan A+\cot A}$

Converting in sin and cos form,

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

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

= $\frac{1}{\frac{1}{\cos A \sin A}}$

= $\cos A \sin A$

So, LHS = RHS

Hence, proved.

Answer 2

Answer:

Step-by-step explanation: