Question

(a) Prove that $\frac{c o t A - c o s A}{c o t A + c o s A} = \frac{s e c A - t a n A}{s e c A + t a n A}$
OR
(b) Prove that $(csc A - sin A) (sec A - cos A) = \frac{1}{t a n A + c o t A}$

Approach

Prove the trigonometric identity $\frac{c o t A - c o s A}{c o t A + c o s A} = \frac{s e c A - t a n A}{s e c A + t a n A}$ by simplifying the LHS. Divide numerator and denominator by $cos A$ , use $1 - sin A = (sec A - tan A) (sec A + tan A)^{- 1}$ form, and simplify to match RHS.

Step-by-step working

1
LHS $= \frac{c o s A - s i n A c o s A}{c o s A + s i n A c o s A}$ (divide numerator and denominator by $sin A$ to get $cot A = \frac{c o s A}{s i n A}$ ). Actually, write $cot A = \frac{c o s A}{s i n A}$ , so LHS $= \frac{\frac{c o s A}{s i n A} - c o s A}{\frac{c o s A}{s i n A} + c o s A} = \frac{c o s A ( \frac{1}{s i n A} - 1 )}{c o s A ( \frac{1}{s i n A} + 1 )} = \frac{1 - s i n A}{1 + s i n A}$ .
1 mark
2
Simplify: $\frac{1 - s i n A}{1 + s i n A}$ . Divide numerator and denominator by $cos A$ : $\frac{\frac{1}{c o s A} - \frac{s i n A}{c o s A}}{\frac{1}{c o s A} + \frac{s i n A}{c o s A}} = \frac{s e c A - t a n A}{s e c A + t a n A}$ = RHS.
1.5 marks
3
Hence LHS = RHS, identity proved.
0.5 marks

Concepts used

trigonometric identityprovesimplify

Approach

Step-by-step working

Concepts used

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