Question

Prove the following identities. (Remember, you may only change one side of the equation)

(a) Sec θ (1 - sin^2 θ) = cos θ (b) 1 = 1 + sin x/ sin x - csc x

Answer 1

Explanation:

`(a)\\\\\sec\theta(1-\sin^2\theta)=\cos\theta\\\\L_S=\sec\theta(1-\sin^2\theta)\\\\\text{use}\ \sec x=(1)/(\cos x)\ \text{and}\ \sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\L_S=(1)/(\cos\theta)\cdot\cos^2\theta=(\cos^2\theta)/(\cos\theta)=\cos\theta=R_S\\\\\blacksquare`

`(b)\\\\1=(1+\sin x)/(\sin x-\csc x)\\\\R_S=(1+\sin x)/(\sin x-\csc x)=(1+\sin x)/(\sin x-\csc x)\\\\\text{use}\ \csc x=(1)/(\sin x)\\\\R_S=(1+\sin x)/\left(\sin x-(1)/(\sin x)\right)=(1+\sin x)/\left((\sin^2x)/(\sin x)-(1)/(\sin x)\right)\\\\R_S=(1+\sin x)/\left((\sin^2x-1)/(\sin x)\right)=(1+\sin x)\cdot\left((\sin x)/(1-\sin^2x)\right)\\\\R_S=(1+\sin x)\cdot\left((\sin x)/(1^2-\sin^2x)\right)\\\\\text{use}\ a^2-b^2=(a-b)(a+b)`

`R_S=(1+\sin x)\cdot\left((\sin x)/((\sin x-1)(\sin x+1))\right)\\\\R_S=((1+\sin x)(\sin x))/((\sin x-1)(\sin x+1))\\\\\text{cancel}\ (\sin x+1)\\\\R_S=(\sin x)/(\sin x-1)=(\sin x-1+1)/(\sin x-1)=(\sin x-1)/(\sin x-1)+(1)/(\sin x-1)\\\\R_S=1+(1)/(\sin x-1)\\\\\text{If there is to be equality} \ R_S=L_S,\ \text{then}\\\\1=1+(1)/(\sin x-1)\qquad\text{subtract 1 from both sides}\\\\0=(1)/(\sin x-1)\to\bold{it's\ impossible\ for\ each\ values\ of\ }x`

`\text{CONCLUSION:}\\\\\huge{\text{It's not an identity}}`