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# Trigonometric Identity?

(cos^2x) / (1+2sin(x)-3sin^2(x)) = (1 + sin(x)) / (1 + 3sin(x))

Please show ALL the steps involved so that I can understand it better.

Thank you in advance.

### 2 Answers

- Scrander berryLv 71 decade agoFavorite Answer
LHS = cos²(x) / [1 + 2sin(x) - 3sin²(x)]

= [1 - sin²(x)] / [(1 + 3sin(x))(1 - sin(x))]

= [(1 - sin(x))(1 + sin(x))] / [(1 + 3sin(x))(1 - sin(x))]

= [1 + sin(x)] / [1 + 3sin(x)]

= RHS qed

- 1 decade ago
i am working on the left hand side

numerator

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

1-sin^2(x)=(1-sin(x))(1+sin(x)) (difference of squares)

the denominator is a trigonometric quadratic equation

so factor it

(1+2sin(x)-3sin^2(x))= -(3sin^2(x)-2sin(x)-1)

-(3sin^2(x)-2sin(x)-1)=-(3sin^2(x)-3sin(x)+sin(x)-1)

=-( 3sin(x)(sin(x)-1)+1(sin(x)-1))

=-(3sin(x)+1)(sin(x)-1)

multiply the second bracket by the negative sign

=(3sin(x)+1)(1-sin(x))

now the left hand side becomes

(1-sin(x))(1+sin(x))/(3sin(x)+1)(1-sin(x))

=(1+sin(x))/(3sin(x)+1)

there fore left hand side = right hand side