Hi! It will be a pleasure to help you to prove these identities, so let's get started:
PART a)
We have the following expression:
[tex]tan(\theta)cot(\theta)-sin^{2}(\theta)=cos^2(\theta)[/tex]
We know that:
[tex]cot(\theta)=\frac{1}{cot(\theta)}[/tex]
Therefore, by substituting in the original expression:
[tex]tan(\theta)\left(\frac{1}{tan(\theta)}\right)-sin^{2}(\theta)=cos^2(\theta) \\ \\ \\ Simplifying: \\ \\ 1-sin^2(\theta)=cos^2(\theta)[/tex]
We know that the basic relationship between the sine and the cosine determined by the Pythagorean identity, so:
[tex]sin^2(\theta)+cos^2(\theta)=1[/tex]
By subtracting [tex]sin^2(\theta)[/tex] from both sides, we get:
[tex]\boxed{cos^2(\theta)=1-sin^2(\theta)} \ Proved![/tex]
PART b)
We have the following expression:
[tex]\frac{cos(\alpha)}{cos(\alpha)-sin(\alpha)}=\frac{1}{1-tan(\alpha)}[/tex]
Here, let's multiply each side by [tex]cos(\alpha)-sin(\alpha)[/tex]:
[tex](cos(\alpha)-sin(\alpha))\left(\frac{cos(\alpha)}{cos(\alpha)-sin(\alpha)}\right)=(cos(\alpha)-sin(\alpha))\left(\frac{1}{1-tan(\alpha)}\right) \\ \\ Then: \\ \\ cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{1-tan(\alpha)}[/tex]
We also know that:
[tex]tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}[/tex]
Then:
[tex]cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{1-\frac{sin(\alpha)}{cos(\alpha)}} \\ \\ \\ Simplifying: \\ \\ cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)}} \\ \\ Or: \\ \\ cos(\alpha)=\frac{\frac{cos(\alpha)-sin(\alpha)}{1}}{\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)}} \\ \\ Then: \\ \\ cos(\alpha)=cos(\alpha).\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)-sin(\alpha)} \\ \\ \boxed{cos(\alpha)=cos(\alpha)} \ Proved![/tex]
PART c)
We have the following expression:
[tex]\frac{cos(x+y)}{cosxsiny}=coty-tanx[/tex]
From Angle Sum Property, we know that:
[tex]cos(x+y)=cos(x)cos(y)-sin(x)sin(y)[/tex]
Substituting this in our original expression, we have:
[tex]\frac{cos(x)cos(y)-sin(x)sin(y)}{cosxsiny}=coty-tanx[/tex]
But we can also write this as follows:
[tex]\\ \frac{cosxcosy}{cosxsiny}-\frac{sinxsiny}{cosxsiny}=coty-tanx \\ \\ Simplifying: \\ \\ \frac{cosy}{siny}-\frac{sinx}{cosx} =coty-tanx \\ \\ But: \\ \\ \frac{cosy}{siny}=coty \\ \\ \frac{sinx}{cosx}=tanx \\ \\ Hence: \\ \\ \boxed{coty-tanx=coty-tanx} \ Proved![/tex]
PART d)
We have the following expression:
[tex]\ln\left|1+cos \theta\right|+\ln\left|1-cos \theta\right|=2\ln\left|sin \theta\right|[/tex]
By Logarithm product rule, we know:
[tex]log_{b}(x.y) = log_{b}(x) + log_{b}(y)[/tex]
So:
[tex]\ln\left|1+cos \theta\right|+\ln\left|1-cos \theta\right|=\ln\left|(1+cos \theta)(1-cos \theta)\right|[/tex]
The Difference of Squares states that:
[tex]a^2-b^2=(a+b)(a-b) \\ \\ So: \\ \\ (1+cos \theta)(1-cos \theta)=1-cos^2 \theta[/tex]
Then:
[tex]\ln\left|(1+cos \theta)(1-cos \theta)\right|=\ln\left|1-cos^{2} \theta\right|[/tex]
By the Pythagorean identity:
[tex]sin^2(\theta)+cos^2(\theta)=1 \\ \\ So: \\ \\ sin^2 \theta = 1-cos^2 \theta[/tex]
Then:
[tex]\ln\left|1-cos^{2} \theta\right|=\ln\left|sin^2 \theta|[/tex]
By Logarithm power rule, we know:
[tex]log_{b}(x.y) = ylog_{b}(x)[/tex]
Then:
[tex]\ln\left|sin^2 \theta|=2\ln\left|sin \theta|[/tex]
In conclusion:
[tex]\boxed{\ln\left|1+cos \theta\right|+\ln\left|1-cos \theta\right|=2\ln\left|sin \theta\right|} \ Proved![/tex]
