Step-by-step explanation:
[tex](a)\\\\\sec\theta(1-\sin^2\theta)=\cos\theta\\\\L_S=\sec\theta(1-\sin^2\theta)\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\ \text{and}\ \sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\L_S=\dfrac{1}{\cos\theta}\cdot\cos^2\theta=\dfrac{\cos^2\theta}{\cos\theta}=\cos\theta=R_S\\\\\blacksquare[/tex]
[tex](b)\\\\1=\dfrac{1+\sin x}{\sin x-\csc x}\\\\R_S=\dfrac{1+\sin x}{\sin x-\csc x}=(1+\sin x)\div(\sin x-\csc x)\\\\\text{use}\ \csc x=\dfrac{1}{\sin x}\\\\R_S=(1+\sin x)\div\left(\sin x-\dfrac{1}{\sin x}\right)=(1+\sin x)\div\left(\dfrac{\sin^2x}{\sin x}-\dfrac{1}{\sin x}\right)\\\\R_S=(1+\sin x)\div\left(\dfrac{\sin^2x-1}{\sin x}\right)=(1+\sin x)\cdot\left(\dfrac{\sin x}{1-\sin^2x}\right)\\\\R_S=(1+\sin x)\cdot\left(\dfrac{\sin x}{1^2-\sin^2x}\right)\\\\\text{use}\ a^2-b^2=(a-b)(a+b)[/tex]
[tex]R_S=(1+\sin x)\cdot\left(\dfrac{\sin x}{(\sin x-1)(\sin x+1)}\right)\\\\R_S=\dfrac{(1+\sin x)(\sin x)}{(\sin x-1)(\sin x+1)}\\\\\text{cancel}\ (\sin x+1)\\\\R_S=\dfrac{\sin x}{\sin x-1}=\dfrac{\sin x-1+1}{\sin x-1}=\dfrac{\sin x-1}{\sin x-1}+\dfrac{1}{\sin x-1}\\\\R_S=1+\dfrac{1}{\sin x-1}\\\\\text{If there is to be equality} \ R_S=L_S,\ \text{then}\\\\1=1+\dfrac{1}{\sin x-1}\qquad\text{subtract 1 from both sides}\\\\0=\dfrac{1}{\sin x-1}\to\bold{it's\ impossible\ for\ each\ values\ of\ }x[/tex]
[tex]\text{CONCLUSION:}\\\\\huge{\text{It's not an identity}}[/tex]
