Question

Can someone please help me?

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

Answer 2

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: