Question

Prove that:
$\frac{1 + cot ^{2} A}{1 + tan ^{2} A} = (\frac{1 - cot A}{1 - tan A})^{2}$

Approach

Prove the identity $\frac{1 + c o t ^{2} A}{1 + t a n ^{2} A} = (\frac{1 - c o t A}{1 - t a n A})^{2}$ . Simplify LHS: $\frac{c s c ^{2} A}{s e c ^{2} A} = \frac{c o s ^{2} A}{s i n ^{2} A} = cot^{2} A$ . Simplify RHS: $(\frac{s i n A - c o s A}{c o s A - s i n A})^{2} = (\frac{1 - c o t A}{1 - t a n A})^{2} = cot^{2} A$ . Thus LHS = RHS.

Step-by-step working

1
LHS = $\frac{1 + c o t ^{2} A}{1 + t a n ^{2} A} = \frac{c s c ^{2} A}{s e c ^{2} A} = \frac{c o s ^{2} A}{s i n ^{2} A}$ .
1 mark
2
RHS = $(\frac{s i n A - c o s A}{s i n A} \div \frac{c o s A - s i n A}{c o s A})^{2} = (\frac{s i n A - c o s A}{c o s A - s i n A})^{2} = (- \frac{c o s A}{s i n A})^{2}$ .
1.5 marks
3
RHS = $\frac{c o s ^{2} A}{s i n ^{2} A} =$ LHS.
0.5 marks

Concepts used

trigonometric identityprovecot squaredtan squaredsimplify

Approach

Step-by-step working

Concepts used

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