1.

Prove that : (secA + tan A)(1– sinA) = cosA.

Answer»

. L.H.S = (secA + tanA) (1 – sinA) 

= (\(\frac{1}{cosA}+\frac{sinA}{cosA}\))(1 – sinA) (∵ secA = \(\frac{1}{cos A}\) & tanA = \(\frac{sinA}{cosA}\)

= ( \(\frac{1+sin^2A}{cosA}\) ) (1 – sinA) 

= (\(\frac{1 − sin^2A }{cosA}\)) [∵ (a + b)(a − b) = a2 − b2

= \(\frac{cos^2A}{cosA}\) = cosA = R.H.S. (∵ sin2A + cos2A = 1 ⇒ 1 − sin2A = cos2A) 

Hence proved



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