Question

1 + cot A + tan a into Sin A minus Cos A is equal to sec a upon cos square A minus Cos A upon sec square A​

Answer 1

this is the answer please like it

Answer 2

To prove: $(1+\cot A +\tan A)(\sin A -\cos A)=\dfrac{\sec A}{\text {cosec}^2 A}- \dfrac{\text {cosec}}{\sec ^2 A}$

Step-by-step explanation:

Consider L.H.S.

$(1+\cot A +\tan A)(\sin A -\cos A)\\\\= (1 + \dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A})(\sin A- \cos A)\\\\= (1+ \dfrac{\cos^2 A +\sin ^2 A}{\sin A \cos A} )(\sin A- \cos A)\\\\=(1+\dfrac{1}{\sin A \cos A}) (\sin A- \cos A) \tex{------ }(\because \sin^2A +\cos^2A =1})\\\\= \dfrac{1+\sin A \cos A}{\sin A \cos A} (\sin A- \cos A)$

Now consider R.H.S.

we have

$\dfrac{\sec A}{\text {cosec}^2 A}- \dfrac{\text {cosec}}{\sec ^2 A}\\\\=\dfrac{\sin^2 A}{\cos A} -\dfrac{\cos^2 A}{\sin A} \\\\= \dfrac{\sin^3 A-\cos ^3 A}{\sin A\cos A} \\\\= \dfrac{(\sin A-\cos A)(\sin ^2 A +\cos ^2 A + \sin A\cos A)}{\sin A \cos A} \\\\\text{}[\because a^3-b^3=(a-b)(a^2+b^2+ab)]\\\\=\dfrac{(\sin A-\cos A)(1 + \sin A\cos A)}{\sin A \cos A}$

Now as L.H.S. = R.H.S

Hence, proved the required result