Answer:
option C Only second equation is an identity is correct.
Step-by-step explanation:
1)
[tex]1 +\frac{cos^2\theta}{cot^2\theta(1-sin^2\theta)}= 9 sec^2\theta[/tex]
We need to prove this identity.
We know:
[tex]cos^2\theta + sin^2\theta = 1\\=> cos^2\theta = 1- sin^2\theta[/tex]
and
[tex]\frac{1}{cot^2\theta } = tan^2\theta \\and\\tan^2\theta = \frac{sin^2\theta}{cos^2\theta}[/tex]
Using these to solve the identity
[tex]1 +\frac{cos^2\theta}{cot^2\theta(cos^2\theta)}= 9 sec^2\theta\\1 +\frac{1}{cot^2\theta} = 9 sec^2\theta\\1+tan^2\theta = 9 sec^2\theta\\1+\frac{sin^2\theta}{cos^2\theta} = 9sec^2\theta\\\\\frac{cos^2\theta+sin^2\theta}{cos^2\theta} = 9sec^2\theta\\\frac{1}{cos^2\theta}=9sec^2\theta \\sec^2\theta \neq 9sec^2\theta[/tex]
So, this is not an identity.
2)
[tex]20sin\theta(\frac{1}{sin\theta} -\frac{cot\theta}{sec\theta}) =20sin^2\theta\\[/tex]
We need to prove this identity.
We know:
[tex]cot\theta = \frac{cos\theta}{sin\theta} \\and \\sec\theta=\frac{1}{cos\theta} \\so,\,\, \frac{cot\theta}{sec\theta}= \frac{\frac{cos\theta}{sin\theta}}{\frac{1}{cos\theta}} \\Solving\\ \frac{cot\theta}{sec\theta} =\frac{cos^2\theta}{sin\theta}[/tex]
Using this to solve the identity
[tex]20sin\theta(\frac{1}{sin\theta} -\frac{cot\theta}{sec\theta}) =20sin^2\theta\\\\Putting\,\,values\,\,\\ 20sin\theta(\frac{1}{sin\theta} -\frac{cos^2\theta}{sin\theta}) =20sin^2\theta\\ 20sin\theta(\frac{1-cos^2\theta}{sin\theta}) =20sin^2\theta\\1-cos^2\theta = sin^2\theta\\ 20sin\theta(\frac{sin^2\theta}{sin\theta}) =20sin^2\theta\\Cancelling\,\, sin\theta \,\,over \,\,sin\theta\\20sin^2\theta=20sin^2\theta[/tex]
So, this is an identity.
So, option C Only second equation is an identity is correct.
