Question

(sec a-tan a )^2(1+sin a )=1 -sina
Prove that lhs = rhs

Answer 1

LHS

(sec a - tan a)² ( 1+sin a)

= (sec²a + tan²a -2sec a tan a) (1+sin a)

= (1+tan²a+tan²a -2sec a tan a)(1+sin a)

= (1+2tan²a -2sec a tan a)(1+sin a)

tan a = sin a/cos a

= (cos²a + 2sin²a -2sin a)(1+sin a)/cos²a

= (1-sin²a+2sin² a -2sin a)(1+sina)/(1-sin²a)

= (1+sin²a-2sina)(1+sina) / (1-sin a)(1+sin a)

= (1-sin a)² (1+sin a)/(1-sin a)(1+sin a)

= 1-sin a

= RHS

Answer 2

LHS

LHS= (sec A - tan A)²; express all as sine and cosine

LHS= (sec A - tan A)²; express all as sine and cosine= (1/cos A - sin A/cos A)²

LHS= (sec A - tan A)²; express all as sine and cosine= (1/cos A - sin A/cos A)²= [(1 - sin A)/cos A)]²

LHS= (sec A - tan A)²; express all as sine and cosine= (1/cos A - sin A/cos A)²= [(1 - sin A)/cos A)]²= (1 - sin A)²/cos² A

LHS= (sec A - tan A)²; express all as sine and cosine= (1/cos A - sin A/cos A)²= [(1 - sin A)/cos A)]²= (1 - sin A)²/cos² A= (1 - sin A)²/(1 - sin² A)

LHS= (sec A - tan A)²; express all as sine and cosine= (1/cos A - sin A/cos A)²= [(1 - sin A)/cos A)]²= (1 - sin A)²/cos² A= (1 - sin A)²/(1 - sin² A)= (1 - sin A)²/[(1 - sin A)(1 + sin A)

LHS= (sec A - tan A)²; express all as sine and cosine= (1/cos A - sin A/cos A)²= [(1 - sin A)/cos A)]²= (1 - sin A)²/cos² A= (1 - sin A)²/(1 - sin² A)= (1 - sin A)²/[(1 - sin A)(1 + sin A)= (1 - sin A)/(1 + sin A)

LHS= (sec A - tan A)²; express all as sine and cosine= (1/cos A - sin A/cos A)²= [(1 - sin A)/cos A)]²= (1 - sin A)²/cos² A= (1 - sin A)²/(1 - sin² A)= (1 - sin A)²/[(1 - sin A)(1 + sin A)= (1 - sin A)/(1 + sin A)= RHS

OR

RHS

RHS= (1-sinA)/(1+sinA)

RHS= (1-sinA)/(1+sinA)= (1-sinA)/(1+sinA) × (1-sinA)/(1-sinA) By Rationalization

RHS= (1-sinA)/(1+sinA)= (1-sinA)/(1+sinA) × (1-sinA)/(1-sinA) By Rationalization= (1-sinA)²/(1+sinA)(1-sinA)

RHS= (1-sinA)/(1+sinA)= (1-sinA)/(1+sinA) × (1-sinA)/(1-sinA) By Rationalization= (1-sinA)²/(1+sinA)(1-sinA)= (1-sinA)/(1+sinA) = LHS