Question

cotA cot4A +1
cotA cot4A-1
cos3A
cos5A
​

Answer 1

Answer:

Step-by-step explanation:

We have to prove that:

\frac{\cot A \cot 4A +1}{\cot A \cot 4A-1}= \frac{\cos 3A}{\cos 5A}

Consider LHS,

\frac{\cot A \cot 4A +1}{\cot A \cot 4A-1}

= \frac{\frac{\cos A}{\sin A}\frac{\cos (4A)}{\sin (4A)}+ \frac{\sin (4A)}{\sin (4A)}\frac{\sin A}{\sin A}}{\frac{\cos A}{\sin A}\frac{\cos (4A)}{\sin (4A)}- \frac{\sin (4A)}{\sin (4A)}\frac{\sin A}{\sin A}}

= \frac{\frac{\cos A \cos (4A)+ \sin (4A) \sin A}{\sin A \sin(4A)}}{\frac{\cos A \cos (4A)- \sin (4A) \sin A}{\sin A \sin(4A)}}

= \frac{\cos A \cos (4A)+ \sin (4A) \sin A}{\cos A \cos (4A)- \sin (4A) \sin A}

Using trigonometric identities

\cos(A-B) = \cos A \cos B + \sin A \sin B and

\cos(A+B) = \cos A \cos B - \sin A \sin B

So, we get

= \frac{\cos(A-4A)}{\cos(A+4A)}

= \frac{\cos(-3A)}{\cos(5A)}

= \frac{\cos(3A)}{\cos(5A)}

= RHS

Hence, proved