Answer:
option C Only second equation is an identity is correct.
Step-by-step explanation:
1)
![1 +\frac{cos^2\theta}{cot^2\theta(1-sin^2\theta)}= 9 sec^2\theta](https://tex.z-dn.net/?f=1%20%2B%5Cfrac%7Bcos%5E2%5Ctheta%7D%7Bcot%5E2%5Ctheta%281-sin%5E2%5Ctheta%29%7D%3D%209%20sec%5E2%5Ctheta)
We need to prove this identity.
We know:
![cos^2\theta + sin^2\theta = 1\\=> cos^2\theta = 1- sin^2\theta](https://tex.z-dn.net/?f=cos%5E2%5Ctheta%20%2B%20sin%5E2%5Ctheta%20%3D%201%5C%5C%3D%3E%20cos%5E2%5Ctheta%20%3D%201-%20sin%5E2%5Ctheta)
and
![\frac{1}{cot^2\theta } = tan^2\theta \\and\\tan^2\theta = \frac{sin^2\theta}{cos^2\theta}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bcot%5E2%5Ctheta%20%7D%20%3D%20tan%5E2%5Ctheta%20%5C%5Cand%5C%5Ctan%5E2%5Ctheta%20%3D%20%5Cfrac%7Bsin%5E2%5Ctheta%7D%7Bcos%5E2%5Ctheta%7D)
Using these to solve the identity
![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](https://tex.z-dn.net/?f=1%20%2B%5Cfrac%7Bcos%5E2%5Ctheta%7D%7Bcot%5E2%5Ctheta%28cos%5E2%5Ctheta%29%7D%3D%209%20sec%5E2%5Ctheta%5C%5C1%20%2B%5Cfrac%7B1%7D%7Bcot%5E2%5Ctheta%7D%20%3D%209%20sec%5E2%5Ctheta%5C%5C1%2Btan%5E2%5Ctheta%20%3D%209%20sec%5E2%5Ctheta%5C%5C1%2B%5Cfrac%7Bsin%5E2%5Ctheta%7D%7Bcos%5E2%5Ctheta%7D%20%3D%209sec%5E2%5Ctheta%5C%5C%5C%5C%5Cfrac%7Bcos%5E2%5Ctheta%2Bsin%5E2%5Ctheta%7D%7Bcos%5E2%5Ctheta%7D%20%3D%209sec%5E2%5Ctheta%5C%5C%5Cfrac%7B1%7D%7Bcos%5E2%5Ctheta%7D%3D9sec%5E2%5Ctheta%20%5C%5Csec%5E2%5Ctheta%20%5Cneq%209sec%5E2%5Ctheta)
So, this is not an identity.
2)
![20sin\theta(\frac{1}{sin\theta} -\frac{cot\theta}{sec\theta}) =20sin^2\theta\\](https://tex.z-dn.net/?f=20sin%5Ctheta%28%5Cfrac%7B1%7D%7Bsin%5Ctheta%7D%20-%5Cfrac%7Bcot%5Ctheta%7D%7Bsec%5Ctheta%7D%29%20%3D20sin%5E2%5Ctheta%5C%5C)
We need to prove this identity.
We know:
![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}](https://tex.z-dn.net/?f=cot%5Ctheta%20%3D%20%5Cfrac%7Bcos%5Ctheta%7D%7Bsin%5Ctheta%7D%20%5C%5Cand%20%5C%5Csec%5Ctheta%3D%5Cfrac%7B1%7D%7Bcos%5Ctheta%7D%20%5C%5Cso%2C%5C%2C%5C%2C%20%5Cfrac%7Bcot%5Ctheta%7D%7Bsec%5Ctheta%7D%3D%20%5Cfrac%7B%5Cfrac%7Bcos%5Ctheta%7D%7Bsin%5Ctheta%7D%7D%7B%5Cfrac%7B1%7D%7Bcos%5Ctheta%7D%7D%20%20%5C%5CSolving%5C%5C%20%5Cfrac%7Bcot%5Ctheta%7D%7Bsec%5Ctheta%7D%20%3D%5Cfrac%7Bcos%5E2%5Ctheta%7D%7Bsin%5Ctheta%7D)
Using this to solve the identity
![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](https://tex.z-dn.net/?f=20sin%5Ctheta%28%5Cfrac%7B1%7D%7Bsin%5Ctheta%7D%20-%5Cfrac%7Bcot%5Ctheta%7D%7Bsec%5Ctheta%7D%29%20%3D20sin%5E2%5Ctheta%5C%5C%5C%5CPutting%5C%2C%5C%2Cvalues%5C%2C%5C%2C%5C%5C%2020sin%5Ctheta%28%5Cfrac%7B1%7D%7Bsin%5Ctheta%7D%20-%5Cfrac%7Bcos%5E2%5Ctheta%7D%7Bsin%5Ctheta%7D%29%20%3D20sin%5E2%5Ctheta%5C%5C%2020sin%5Ctheta%28%5Cfrac%7B1-cos%5E2%5Ctheta%7D%7Bsin%5Ctheta%7D%29%20%3D20sin%5E2%5Ctheta%5C%5C1-cos%5E2%5Ctheta%20%3D%20sin%5E2%5Ctheta%5C%5C%2020sin%5Ctheta%28%5Cfrac%7Bsin%5E2%5Ctheta%7D%7Bsin%5Ctheta%7D%29%20%3D20sin%5E2%5Ctheta%5C%5CCancelling%5C%2C%5C%2C%20sin%5Ctheta%20%5C%2C%5C%2Cover%20%5C%2C%5C%2Csin%5Ctheta%5C%5C20sin%5E2%5Ctheta%3D20sin%5E2%5Ctheta)
So, this is an identity.
So, option C Only second equation is an identity is correct.