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 ]

Adding these two fractions,

[ 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:

• cot(θ) = cos(θ) / sin(θ)
• sin²(θ) + cos²(θ) = 1