Question

1)Verify the identity. Show your work.

cot2x + csc2x = 2csc2x - 1


 cot²x + csc²x = 2csc²x − 1


  cos²x   1      2
 = ▬▬▬ + ▬▬▬ = ▬▬▬ − 1
   sin²x      sin²x     sin²x
2)Verify the identity. Show your work.
1 + sec2xsin2x = sec2x
my work plzz check if wrong plz correct and explain

os^2x+sin^2x=1
so (if you divide both sides by cos^2x)...
1+tan^2x=sec^2x tan^2x=sec^2x-1

sec^2x/sec^2x-1=csc^2x
sec^2x/tan^2x=csc^2x
(1/cos^2x)/(sin^2x/cos^2x)=csc^2x
(1/cos^2x)(cos^2x/sin^2x)=csc^2x
(1/sin^2x)=csc^2x
csc^2x=csc^2x

Answer 1

@mathstudent55

Answer 2

@mathmath333 here it is

Answer 3

for Q-1 Nnesha heped me but i am cofused if it was solved?

Answer 4

cot=cos/sin
sin=1/csc csc=3
cos=1/sec sec=1/cos
is it okay?

Answer 5

1st question is this
\[\cot2x + \csc2x = 2\csc2x - 1\]

or this

\(cot^2x + csc^2x = 2csc^2x - 1\)

Answer 6

u should only post question in the box that extra thing is confusing me

Answer 7

oh i am sorry second one is the q u wrote

Answer 8

\(\large\tt \begin{align} \color{black}{
cot^2x + cosec^2x \\~\\
=cosec^2x+cot^2x \\~\\
=2cosec^2x-cosec^2x+cot^2x \\~\\
=2cosec^2x-\dfrac{1}{sin^2x}+\dfrac{cos^2x}{sin^2x} \\~\\
=2cosec^2x+\dfrac{-1+cos^2x}{sin^2x} \\~\\
=2cosec^2x+\dfrac{-1+cos^2x}{sin^2x} \\~\\
=2cosec^2x+\dfrac{-sin^2x}{sin^2x} \\~\\
=2cosec^2x-1 \\~\\
}\end{align}\)

Answer 9

thanks can post few more? @mathmath333

Answer 10

\[\huge 1 + \sec2xsin2x = \sec2x \\or~this\\ \huge1 + \sec^2xsin^2x = \sec^2x \]

Answer 11

second one @mathmath333 (^) this means raise the power of