Question

cosA-sinA/coA+sinA=sec2A-tan2A

Answer 1

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Answer 2

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

Step-by-step explanation:

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

Rationalising$\cos A+\sin A$, we get

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

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

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

[∵ $\cos ^{2} A-\sin^{2} A=\cos 2A$ ]

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

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

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

$=\sec 2A-\tan 2A$

= R.H.S.. proved.