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# Trigonometry Pproblems Help!!!!!! *10 Pointsss**?

Use trigonometric identities to transform the left side of the equation to the right one.

*** I'm using Q as theta***

1) tanQ CosQ= SinQ

2) (1+CosQ)(1-CosQ)=sin^2Q

3) (SecQ+TanQ)(SecQ-TanQ)=1

4) SinQ/CosQ=CscQ SecQ

### 1 Answer

- Asnyc05Lv 61 decade agoFavorite Answer
1.

Starting from the left hand side (LHS):

= tanθ.cosθ

= (sinθ/cosθ).cosθ

Cancel the common cosθ from numerator and denominator:

= sinθ

= RHS...Proved

2.

Starting from the left hand side (LHS):

= (1+cosθ)(1-cosθ)

This is essentially (a+b)(a-b) = a²-b²

= 1-cos²θ

= sin²θ

[sin²θ+cos²θ = 1, this is the Pythagorean identity. Just subtract cos²θ from both sides of this identity]

= RHS...Proved

3.

Starting from the left hand side (LHS):

= (secθ+tanθ)(secθ-tanθ)

= sec²θ-tan²θ [same as in #2]

= 1 [same as in #2, sec²θ-tan²θ = 1]

= RHS...Proved

Alternatively, convert sec and tan to their respective sine/cosine forms, and you will see that they are the same:

= sec²θ-tan²θ

= (1/cos²θ)-(sin²θ/cos²θ)

= (1-sin²θ)/cos²θ

= cos²θ/cos²θ

= 1

4.

Not sure about this one. Is the question correct?

sinθ/cosθ = tanθ and cscθ.secθ = 1/(sinθ.cosθ).

Can't think of a way to prove 1/(sinθ.cosθ) = tanθ.