Any Thoughts?

Question

Hello,

There are some questions that I am not sure how to tackle:

1) sin^4x - cos^4x = sin^2x - cos^2x

2) sec^4x - tan^4x = 1 + 2tan^2x

3) (sinθ) / (1 - cosθ) + (sinθ) / (1 + cosθ) = 2cosecθ

4) (cot^2θ) / (1 + cot^2θ) = cos^2θ

Thank you in advance.

Answer 1

1) sin^4x - cos^4x = sin^2x - cos^2x

Left hand side identity = sin⁴(θ)- cos⁴(θ)

Apply Difference of Two Squares : a²- b² = (a + b)(a - b).

                                   = [sin²(θ) + cos²(θ)] * [sin²(θ) - cos²(θ)]

Apply Pythagorean identity : sin²(u) + cos²(u) = 1.
                                  = [sin²(θ) - cos²(θ)]

                                   = Right hand side identity.

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2) sec^4x - tan^4x = 1 + 2tan^2x

Left hand side identity = sec⁴(θ)- tan⁴(θ)

Apply Difference of Two Squares : a²- b² = (a + b)(a - b).

                                   = [sec²(θ) + tan²(θ)] * [sec²(θ) - tan²(θ)]

Apply Pythagorean identity : sec²(θ) - tan²(θ) = 1.

                                   = [sec²(θ) + tan²(θ)]

                                   = [1+tan²(θ) + tan²(θ)]   
                                  = 1+2tan²(θ)

                                   = Right hand side identity.

Answer 2

3) (sinθ) / (1-cosθ) + (sinθ) / (1+cosθ) = 2cosecθ

Left hand side identity (sinθ) / (1-cosθ) + (sinθ) / (1+cosθ)

= [(sinθ)(1+ cosθ)+(sinθ)(1 - cosθ)]/(1+ cosθ)(1 - cosθ)

=[sinθ + sinθcosθ+sinθ - sinθcosθ]/(1 - cos²θ)

=2sinθ / sin²θ

=2 / sinθ

=2cosecθ

= Right hand side identity

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4) (cot^2θ) / (1 + cot^2θ) = cos^2θ

Left hand side identity (cot²θ) / (1 + cot²θ)

Apply Pythagorean identity : cosec²(θ) - cot²(θ) = 1.

                    =  (cot²θ) / (cosec²θ

              =(cot²θ) (sin²θ

                   =(cos²θ/sin²θ) (sin²θ)

                   =cos²θ

= Right hand side identity