conu1JeaHelp0Chaayka
Answered

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

B1 + sec2x sin2x = sec2x

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

D- tan2x + sec2x = 1

Verified Answer :

HomertheGenius

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

ThaNosBigThiCC

Answer:

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  

Step-by-step explanation:

Related Questions