Question

Prove the following trigonometric identity:
$\frac{1 + sec A}{sec A} = \frac{sin ^{2} A}{1 - cos A}$

Approach

LHS $= \frac{1 + s e c A}{s e c A} = \frac{1}{s e c A} + 1 = cos A + 1$ . RHS $= \frac{s i n ^{2} A}{1 - c o s A} = \frac{1 - c o s ^{2} A}{1 - c o s A} = \frac{( 1 - c o s A ) ( 1 + c o s A )}{1 - c o s A} = 1 + cos A$ . LHS $=$ RHS, identity proved.

Step-by-step working

1
LHS $= \frac{1 + s e c A}{s e c A} = cos A + 1$
1 mark
2
RHS $= \frac{s i n ^{2} A}{1 - c o s A} = \frac{1 - c o s ^{2} A}{1 - c o s A} = \frac{( 1 - c o s A ) ( 1 + c o s A )}{1 - c o s A}$
1 mark
3
RHS $= 1 + cos A =$ LHS, identity proved
1 mark

Concepts used

trigonometric identitiesproofsecsincos

Approach

Step-by-step working

Concepts used

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