Trigonometric identity:

**cot**^{2}A - cos^{2}A = cot^{2}A cos^{2}AIn the above trigonometric identity, the LHS is cot

^{2}A - cos

^{2}A. Is there any trigonometric identity (one of the basic ones) which combines cot and cos functions? No: We have 1 + cot

^{2}(θ) = cosec

^{2}(θ) and sin

^{2}(θ) + cos

^{2}(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:

( cosThe cos^{2}(θ) / sin^{2}(θ) ) - cos^{2}(θ)

^{2}(θ) can be written as ( cos

^{2}(θ) / 1 ) ,

[ cosAdding these two fractions,^{2}(θ) / sin^{2}(θ) ] - [ cos^{2}(θ) / 1 ]

[ cosNotice that in the numerator, cos^{2}(θ) - sin^{2}(θ) cos^{2}(θ) ] / sin^{2}(θ)

^{2}(θ) is common,

[ (cosFrom the identity sin^{2}(θ)) (1 - sin^{2}(θ)) ] / sin^{2}(θ)

^{2}(θ) + cos

^{2}(θ) = 1, it follows that cos

^{2}(θ) = 1 - sin

^{2}(θ). Applying this result in the above equation,

[ (cosIn this expression, grouping one cos^{2}(θ)) (cos^{2}(θ)) ] / sin^{2}(θ)

^{2}(θ) with sin

^{2}(θ), we have

[ cosApplying the identity cot^{2}(θ) / sin^{2}(θ) ] * cos^{2}(θ)

^{2}(θ) = cos

^{2}(θ) / sin

^{2}(θ),

= cot... which is the RHS expression^{2}(θ) cos^{2}(θ)

The following identities were applied:

- cot(θ) = cos(θ) / sin(θ)
- sin
^{2}(θ) + cos^{2}(θ) = 1

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