## Pages

### Trigonometry identities - 5

Trigonometric identity: cot2A - cos2A = cot2A cos2A

In the above trigonometric identity, the LHS is cot2A - cos2A. Is there any trigonometric identity (one of the basic ones) which combines cot and cos functions? No: We have 1 + cot2(θ) = cosec2(θ) and sin2(θ) + cos2(θ) = 1, but both of these do not include both cot and cos functions. So which trigonometric identity will we apply to solve it?

One was to solve trigonometric identities (and one of the simplest one) is to, if permissible, convert all functions to sin(θ) and cos(θ) and solve. Since cot(θ) = cos(θ) / sin(θ), so we can write the LHS as follows:
( cos2(θ) / sin2(θ) ) - cos2(θ)
The cos2(θ) can be written as ( cos2(θ) / 1 ) ,
[ cos2(θ) / sin2(θ) ] - [ cos2(θ) / 1 ]
[ cos2(θ) - sin2(θ) cos2(θ) ] / sin2(θ)
Notice that in the numerator, cos2(θ) is common,
[ (cos2(θ)) (1 - sin2(θ)) ] / sin2(θ)
From the identity sin2(θ) + cos2(θ) = 1, it follows that cos2(θ) = 1 - sin2(θ). Applying this result in the above equation,
[ (cos2(θ)) (cos2(θ)) ] / sin2(θ)
In this expression, grouping one cos2(θ) with sin2(θ), we have
[ cos2(θ) / sin2(θ) ] * cos2(θ)
Applying the identity cot2(θ) = cos2(θ) / sin2(θ),
= cot2(θ) cos2(θ)
... which is the RHS expression

The following identities were applied: