Trig Proofs Please Help?

Question

Hello,

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

1) (sinθ) / (1 + sinθ) = tanθ(secθ - tanθ)

2) cosec^2θ(tan^2θ - sin^2θ) = tan^2θ

3) (secθ) / (tanθ + cotθ) = sinθ

4) (cosθ) / (sinθ + 1) + (sinθ + 1) / (cosθ) = 2secθ

Thank you in advance.

Answer 1

1) (sinθ) / (1 + sinθ) = tanθ(secθ - tanθ)

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

Multiply numarator and denominator by 1 - sinθ

= (sinθ) (1 - sinθ) / (1 + sinθ)(1 - sinθ)

= (sinθ) (1 - sinθ) / (1 - sin²θ)

= (sinθ) (1 - sinθ) / (cos²θ)

= (sinθ/cos θ) ((1 - sinθ)/cosθ)

= (tanθ) ((1/cosθ) - ( sinθ/cosθ))

= (tanθ) (secθ - tanθ)

= Right hand side identity.

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2) cosec²θ(tan²θ - sin²θ) = tan²θ

Left hand side identity cosec²θ(tan²θ - sin²θ)

= cosec²θtan²θ - cosec²θsin²θ

= (1/sin²θ) (sin²θ/cos²θ) - (1/sin²θ)(sin²θ)

= (1/cos²θ) - 1

= sec²θ - 1

= tan²θ

= Right hand side identity.

Answer 2

3) (secθ) / (tanθ + cotθ) = sinθ

Left hand side identity (secθ) / (tanθ + cotθ)

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

=(1 / cosθ) / [(sinθsinθ+cosθcosθ) / (sinθcosθ)]

=(1 / cosθ)(sinθcosθ) / (sin²θ + cos²θ)

=sinθ / 1

= sinθ

= Right hand side identity.

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4) (cosθ) / (sinθ + 1) + (sinθ + 1) / (cosθ) = 2secθ

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

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

=[(cos²θ) +(1+sinθ)²]/(cosθ)(sinθ + 1)

=[cos²θ +1+sin²θ+2sinθ]/(cosθ)(sinθ + 1)

=[cos²θ+sin²θ +1+2sinθ]/(cosθ)(sinθ + 1)

=[1 +1+2sinθ]/(cosθ)(sinθ + 1)

=[2+2sinθ]/(cosθ)(sinθ + 1)

=2[1+sinθ]/(cosθ)(sinθ + 1)

=2/cosθ

=2secθ

= Right hand side identity