Question

prove the following identities

sec 2A + tan 2A =(cos A + sin A)/(cos A - sin A)

Answer 1

i hope this will help you

Answer 2

$\sec 2A + \tan 2A =\dfrac{\cos A + \sin A}{\cos A - \sin A},$, proved.

Step-by-step explanation:

To prove, $\sec 2A + \tan 2A =\dfrac{\cos A + \sin A}{\cos A - \sin A}$.

R.H.S.$=\dfrac{\cos A + \sin A}{\cos A - \sin A}$

Rationaising numerator and denominator, we get

$=\dfrac{\cos A + \sin A}{\cos A - \sin A}\times \dfrac{\cos A + \sin A}{\cos A + \sin A}$

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

[ ∵ $a^{2} -b^{2} =(a+b)(a-b)$]

$=\dfrac{\cos^2 A + \sin^2 A+2\cos A\sin A}{\cos 2A}$

[ ∵$\cos^2 A - \sin^2 A=\cos 2A$]

$=\dfrac{1+\sin 2A}{\cos 2A}$

[ ∵ $\sin 2A=2\cos A\sin A$ and $\sin^2 A+\cos^2 A=1$]

$=\dfrac{1}{\cos 2A} +\dfrac{\sin 2A}{\cos 2A}$

$=\sec 2A + \tan 2A$

=R.H.S., proved.

Hence, $\sec 2A + \tan 2A =\dfrac{\cos A + \sin A}{\cos A - \sin A}$, proved.