Question

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

Approach

LHS = $\frac{1 + c o s A}{c o s A}$ . Multiply numerator and denominator by $(1 - sin A)$ : $= \frac{( 1 + c o s A ) ( 1 - s i n A )}{c o s A ( 1 - s i n A )} = \frac{1 - s i n A + c o s A - s i n A c o s A}{c o s A ( 1 - s i n A )}$ . Simplify using $sin^{2} A + cos^{2} A = 1$ to get RHS.

Step-by-step working

1
LHS = $sin^{2} A + csc^{2} A - 2 sin A csc A + cos^{2} A + sec^{2} A - 2 cos A sec A$
1 mark
2
$= (sin^{2} A + cos^{2} A) + (1 + cot^{2} A) + (1 + tan^{2} A) - 2 - 2$
1 mark
3
$= 1 + cot^{2} A + tan^{2} A + 2 - 4 = tan^{2} A + cot^{2} A - 1$ = RHS
0.5 marks
4
Hence proved
0.5 marks

Concepts used

trigonometric identitiesalgebraic manipulationsin cos identityproof

Approach

Step-by-step working

Concepts used

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