Question

Please help me

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

1 + sec2x sin2x = sec2x

[(sin(x))/(1-cos(x))]+[(sin(x))/(1+cos(x))]=2csc(x)

Answer 1

are you sure of your signs for the first one?

Answer 2

Sorry. Its supposed to look like this.\[1+\sec ^{2}x \sin ^{2}x=\sec ^{2}x\]

Answer 3

Ok that looks right.

Typically with trig, it helps to play around with different identities. The most important one in my opinion is sin^2+cos^2=1. All other identities are derived from this. It should look familiar because its also a variation of the Pythagorean Theorem.

1+sec^2 x sin^2 x=sec^2 x
=1+(1/cos^2 x) sin^2 x = 1/cos^2x
=cos^2 x/cos^2 x + sin^2 x/cos^2 x = 1/cos^2 x
= (cos^2x+sin^2 x)/cos^2 x = 1/cos^2 x

multiply both sides by cos^2 x

cos^2x + sin^2 x = 1

Answer 4

For the second problem you can also use the identity sin^2 x+ cos^2 x = 1

(sin(x)/(1-cos(x))+(sin(x)/(1+cos(x)) = 2 csc(x)

multiply the first term (numerator and denominator) by its conjugate 1+cos(x)
multiply the second term (numerator and denominator) by its conjugate 1-cos(x)
csc(x) is just 1/sin(x)

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

you know that sin^2 x + cos^2 x=1
then 1-cos^2 x is just sin^2 x

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

cancel out the sin x in the numerator and one of the sin x from the denominator

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

multiply the left and right side of the equation by sin x to get rid of the sin x

=(1+cos x)+(1-cos x)
=1+cos x+1-cos x = 2
cos x - cos x cancels out

1+1 = 2