Prove the following trigonometric identities:(secA-tanA)2 = \({\frac

1.	Prove the following trigonometric identities:(secA-tanA)2 = \({\frac{1-sinA}{1+sinA}}\)
Answer» (sec A - tan A)²=\(\Big(\frac{1}{cosA}-\frac{sinA}{cosA}\Big)^2\) = \(\frac{(1-sinA)^2}{cos^2A}\) = \(\frac{(1-sinA)^2}{1-sin^2A}\) = \(\frac{(1-sinA)^2}{(1-sinA)(1+sinA)}\) = \(\frac{1-sinA}{1+sinA}\) Hence Proved. sec A = 1/cos A tan A = sin A/ cos A Thus, L.H.S=(1/cosA - sin A/cos A)² =(1-sin A)²/cos² A =(1-sin A)(1- sin A)/(1-sin²A) By a²-b²=(a+b)(a-b) and sin² A + cos²A=1 =(1-sin A)/(1+sin A)=R.H.S(corrected) Hence, proved

Prove the following trigonometric identities:(secA-tanA)2 = \({\frac{1-sinA}{1+sinA}}\)

Answer»

(sec A - tan A)²=\(\Big(\frac{1}{cosA}-\frac{sinA}{cosA}\Big)^2\)

= \(\frac{(1-sinA)^2}{cos^2A}\)

= \(\frac{(1-sinA)^2}{1-sin^2A}\)

= \(\frac{(1-sinA)^2}{(1-sinA)(1+sinA)}\)

= \(\frac{1-sinA}{1+sinA}\)

Hence Proved.

sec A = 1/cos A

tan A = sin A/ cos A

Thus,

L.H.S=(1/cosA - sin A/cos A)²

=(1-sin A)²/cos² A

=(1-sin A)(1- sin A)/(1-sin²A) By a²-b²=(a+b)(a-b) and sin² A + cos²A=1

=(1-sin A)/(1+sin A)=R.H.S(corrected)

Hence, proved