Question

Prove that: $\frac{c o t A - 1}{2 - s e c ^{2} A} = \frac{c o t A}{1 + t a n A}$

Approach

Rewrite LHS using $sec^{2} A = 1 + tan^{2} A$ , so denominator $2 - sec^{2} A = 1 - tan^{2} A$ . Factor as $(1 - tan A) (1 + tan A)$ . Numerator $cot A - 1 = \frac{1 - t a n A}{t a n A}$ . Simplify to get $\frac{c o t A}{1 + t a n A}$ = RHS.

Step-by-step working

1
Express denominator as $2 - sec^{2} A = 1 - tan^{2} A = (1 - tan A) (1 + tan A)$
1 mark
2
Express numerator $cot A - 1 = \frac{1 - t a n A}{t a n A}$
1 mark
3
Cancel and simplify to obtain RHS $\frac{c o t A}{1 + t a n A}$
1 mark

Concepts used

trigonometric identitiesprovesimplificationcot tan sec

Approach

Step-by-step working

Concepts used

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