Question

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

Approach

Prove $\frac{1}{s e c x - t a n x} - \frac{1}{c o s x} = \frac{1}{c o s x} - \frac{1}{s e c x + t a n x}$ . LHS: simplify using $sec^{2} x - tan^{2} x = 1$ to get $sec x + tan x - sec x = tan x$ . RHS: similarly get $tan x$ . Both sides equal.

Step-by-step working

1
LHS = $\frac{1}{s e c x - t a n x} - sec x = \frac{s e c ^{2} x - t a n ^{2} x}{s e c x - t a n x} - sec x = sec x + tan x - sec x = tan x$ .
0.5 marks
2
Simplify: LHS = $tan x$ .
1 mark
3
RHS = $sec x - \frac{s e c ^{2} x - t a n ^{2} x}{s e c x + t a n x} = sec x - (sec x - tan x) = tan x$ .
0.5 marks
4
LHS = RHS. Hence proved.
1 mark

Concepts used

trigonometric identityprovesectancossimplify

Approach

Step-by-step working

Concepts used

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