Question #162013

Trigonometric identities


Solve and prove trigonometric identities


1.) Cot(theta)sin(theta)cos(theta)=1-sin²(theta)


2.) Sin²(theta)sec²(theta)-1


3.) Tan(theta)+cot(theta)= csc(theta)sec(theta)


4.) 1+cot(theta)= csc(theta)(cos(theta) + sin(theta))




1
Expert's answer
2021-02-10T01:10:51-0500

1.)cot(θ)sin(θ)cos(θ)=1sin2(θ)1.)\cot(\theta)\sin(\theta)\cos(\theta)=1-\sin^2(\theta)

cos(θ)sin(θ)(θ)sin(θ)cos(θ)=1sin2(θ)\frac{cos(\theta)}{sin(\theta)}(\theta)\sin(\theta)\cos(\theta)=1-\sin^2(\theta)

cos2(θ)=1sin2(θ)\cos^2(\theta)= 1-\sin^2(\theta)

cos2(θ)+sin2(θ)=1\cos^2(\theta)+\sin^2(\theta)=1

1=1 1=1\ \blacksquare


2.)sin2(θ)sec2(θ)1=2.)\sin^2(\theta)\sec²(\theta)-1=

sin2(θ)1cos2(θ)1=\sin^2(\theta)\frac{1}{cos²(\theta)}-1=

1cos2(θ)cos2(θ)1=1cos2(θ)2=\frac{1-\cos^2(\theta)}{\cos^2(\theta)}-1=\frac{1}{\cos^2(\theta)}-2=

sec2(θ)2\sec^2{(\theta)}-2


3.)tan(θ)+cot(θ)=csc(θ)sec(θ)3.)\tan(\theta)+\cot(\theta)= \csc(\theta)\sec(\theta)

sin(θ)cos(θ)+cos(θ)sin(θ)=csc(θ)sec(θ)\frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}= \csc(\theta)\sec(\theta)

sin2(θ)+sin2(θ)cos(θ)sin(θ)=csc(θ)sec(θ)\frac{\sin^2(\theta)+sin^2(\theta)}{\cos(\theta)\sin(\theta)}= \csc(\theta)\sec(\theta)

1cos(θ)sin(θ)=csc(θ)sec(θ)\frac{1}{\cos(\theta)\sin(\theta)}= \csc(\theta)\sec(\theta)

1cos(θ)1sin(θ)=csc(θ)sec(θ)\frac{1}{\cos(\theta)}\frac{1}{\sin(\theta)}= \csc(\theta)\sec(\theta)

sec(θ)csc(θ)=csc(θ)sec(θ)\sec(\theta)\csc(\theta)=\csc(\theta)\sec(\theta)

csc(θ)sec(θ)=csc(θ)sec(θ) \csc(\theta)\sec(\theta)=\csc(\theta)\sec(\theta)\ \blacksquare


4.)1+cot(θ)=csc(θ)(cos(θ)+sin(θ))4.)1+\cot(\theta)= \csc(\theta)(cos(\theta) + sin(\theta))

1+cot(θ)=1sin(θ)(cos(θ)+sin(θ))1+\cot(\theta)= \frac{1}{sin(\theta)}(cos(\theta) + sin(\theta))

1+cot(θ)=cos(θ)sin(θ)+11+\cot(\theta)= \frac{\cos(\theta)}{\sin(\theta)}+1

1+cot(θ)=cot(θ)+11+\cot(\theta)= \cot(\theta)+1

1+cot(θ)=1+cot(θ) 1+\cot(\theta)=1+\cot(\theta)\ \blacksquare






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