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idllotsaroms
Feb 6, 2011, 02:15 PM
tan(a-b) = (1-cot(a)tan(b)/cot(a)+tan(b))

idllotsaroms
Feb 6, 2011, 02:16 PM
(cotx)(secx)(sinx)=2-(tanx)(cosx)(cscx)

idllotsaroms
Feb 6, 2011, 05:08 PM
sec(x)cos(x)=2-sin(x)csc(x) is that right

idllotsaroms
Feb 6, 2011, 05:57 PM
(1)/(1+tan^(2)x)+(1)/(1+cot^(2)x)=1

Replace 1+tan^(2)x with sec^(2)x using the identity sec^(2)(x)=1+tan^(2)(x).
(1)/(sec^(2)x)+(1)/(1+cot^(2)x)

Replace 1+cot^(2)x with csc^(2)x using the identity csc^(2)(x)=1+cot^(2)(x).
(1)/(sec^(2)x)+(1)/(csc^(2)x)

To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is secx^(2)cscx^(2). Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
(1)/(sec^(2)x)*(csc^(2)x)/(csc^(2)x)+(1)/(csc^(2)x)*(sec^(2)x)/(sec^(2)x)

Complete the multiplication to produce a denominator of secx^(2)cscx^(2) in each expression.
(csc^(2)x)/(sec^(2)xcsc^(2)x)+(sec^(2)x)/(sec^(2)xcsc^(2)x)

Combine the numerators of all expressions that have common denominators.
(csc^(2)x+sec^(2)x)/(sec^(2)xcsc^(2)x)

Reduce the expression ((cscx^(2)+secx^(2)))/(secx^(2)cscx^(2)) by removing a factor of from the numerator and denominator.
(csc^(2)x+sec^(2)x)/(csc^(2)xsec^(2)x)

Replace cscx with an equivalent expression (1)/(sin^(2)(x)) using the fundamental identities.
(((1)/(sin^(2)(x))+sec^(2)x))/(csc^(2)xsec^(2)x)

Replace secx with an equivalent expression (1)/(cos^(2)(x)) using the fundamental identities.
(((1)/(sin^(2)(x))+(1)/(cos^(2)(x))))/(csc^(2)xsec^(2)x)

To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is sin(x)^(2)cos(x)^(2). Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
((((1)/(sin^(2)(x))*(cos^(2)(x))/(cos^(2)(x))+(1)/(cos^(2)(x))*(sin^(2)(x))/(sin^(2)(x)))))/(csc^(2)xsec^(2)x)

Complete the multiplication to produce a denominator of sin(x)^(2)cos(x)^(2) in each expression.
((((cos^(2)(x))/(sin^(2)(x)cos^(2)(x))+(sin^(2)(x))/(sin^(2)(x)cos^(2)(x)))))/(csc^(2)xsec^(2)x)

Combine the numerators of all expressions that have common denominators.
((((cos^(2)(x)+sin^(2)(x))/(sin^(2)(x)cos^(2)(x)))))/(csc^(2)xsec^(2)x)

Replace sin^(2)(x)+cos^(2)(x) with 1 using the identity sin^(2)(x)+cos^(2)(x)=1.
((((1)/(sin^(2)(x)cos^(2)(x)))))/(csc^(2)xsec^(2)x)

Remove the parentheses from the numerator.
(((1)/(sin^(2)(x)cos^(2)(x))))/(csc^(2)xsec^(2)x)

Remove the parentheses around the expression (1)/(sin(x)^(2)cos(x)^(2)).
((1)/(sin^(2)(x)cos^(2)(x)))/(csc^(2)xsec^(2)x)

Convert the csc^(2)x to its (sin)/(cos) equivalent.
((1)/(sin^(2)(x)cos^(2)(x)))/((((1)/(sin^(2)(x)))))

Convert the sec^(2)x to its (sin)/(cos) equivalent.
((1)/(sin^(2)(x)cos^(2)(x)))/((((1)/(sin^(2)(x)))((1)/(cos^(2)(x)))))

Replace sin(x) with an equivalent expression in the numerator.
((1)/(sin^(2)(x)cos^(2)(x)))/(((csc^(2)(x))((1)/(cos^(2)(x)))))

Replace cos(x) with an equivalent expression in the numerator.
((1)/(sin^(2)(x)cos^(2)(x)))/((csc^(2)(x))(sec^(2)(x)))

Remove the extra parentheses around the factors.
((1)/(sin^(2)(x)cos^(2)(x)))/(csc^(2)(x)(sec^(2)(x)))

Multiply csc(x)^(2) by each term inside the parentheses.
((1)/(sin^(2)(x)cos^(2)(x)))/(csc^(2)(x)sec^(2)(x))

Replace the expressions with an equivalent expression using the fundamental identities.
((1)/(sin^(2)(x)cos^(2)(x)))/(((1)/(sin^(2)(x)cos^(2)(x))))

To divide by (1)/(sin(x)^(2)cos(x)^(2)), multiply by the reciprocal of the fraction.
sin^(2)(x)cos^(2)(x)*(1)/(sin^(2)(x)cos^(2)(x))

Remove the common factor of sin(x)^(2)cos(x)^(2) from the first term sin(x)^(2)cos(x)^(2) and the denominator of the second term (1)/(sin(x)^(2)cos(x)^(2)).
1*1

Multiply 1 by 1 to get 1.
1

idllotsaroms
Feb 6, 2011, 05:58 PM
Is that right

idllotsaroms
Feb 6, 2011, 05:58 PM
NO!

idllotsaroms
Feb 6, 2011, 05:58 PM
Possibly

idllotsaroms
Feb 6, 2011, 06:03 PM
I'm not sure

idllotsaroms
Feb 6, 2011, 06:03 PM
Well everyone on this site is useless so this is the best u get

idllotsaroms
Feb 6, 2011, 06:03 PM
Okay