Question

Can someone 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.

A. (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

B. 1 + sec2x sin2x = sec2x

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

D. - tan2x + sec2x = 1

Answer 1

a ) sin x ( \frac{sinx}{cosx}*cosx- \frac{cosx}{sinx}+*cos x)= \\ sinx(sinx- \frac{cos^{2} x}{sin^{2} x})=
sin^{2}x-cos^{2} x=1-cos^{2}x-cos^{2} x= \\ 1-2cos^{2}x

b ) 1 + \frac{1}{cos^{2} x} *sin^{2} x=1+ \frac{sin^{2} x}{cos^{2} x} = \\ \frac{cos {2} x+sin x^{2} x}{cos^{2} x} = \frac{1}{cos^{2} x} =sec^{2} x

c) \frac{sinx}{1-cosx} + \frac{sinx}{1+cosx}= \frac{sin x ( 1+cosx)+sinx(1-cosx)}{1-cos^{2} x} = \\ \frac{sinx+sinxcosx+sinx-sinxcosxx}{1-cos^{2}x }= \\ \frac{2sinx}{sin^{2} x} = \frac{2}{sinx}=2 csc x

d) -tan ^{2}x+sec ^{2}x=- \frac{sin ^{2} x}{cos ^{2} x} + \frac{1}{cos^{2} x} = \\ \frac{1-sin^{2x} }{cos ^{2}x }= \frac{cos^{2} x}{cos^{2}x } =1

Explanation:

Answer 2

a )
`sin x ( (sinx)/(cosx)*cosx- (cosx)/(sinx)+*cos x)= \\ sinx(sinx- (cos^(2) x)/(sin^(2) x))= sin^(2)x-cos^(2) x=1-cos^(2)x-cos^(2) x= \\ 1-2cos^(2)x`
b )
`1 + (1)/(cos^(2) x) *sin^(2) x=1+ (sin^(2) x)/(cos^(2) x) = \\ \frac{cos {2} x+sin x^(2) x}{cos^(2) x} = (1)/(cos^(2) x) =sec^(2) x`
c)
`(sinx)/(1-cosx) + (sinx)/(1+cosx)= (sin x ( 1+cosx)+sinx(1-cosx))/(1-cos^(2) x) = \\ (sinx+sinxcosx+sinx-sinxcosxx)/(1-cos^(2)x )= \\ (2sinx)/(sin^(2) x) = (2)/(sinx)=2 csc x`
d)
`-tan ^(2)x+sec ^(2)x=- (sin ^(2) x)/(cos ^(2) x) + (1)/(cos^(2) x) = \\ (1-sin^(2x) )/(cos ^(2)x )= (cos^(2) x)/(cos^(2)x ) =1`