Question

Having trouble with the following questions, help would be appreciated.

Answer 1

`\begin{gathered} a)sin\mleft((\pi)/(2)+x\mright)\cdot\: tan\mleft(x\mright) \\ It\text{ is important to know that} \\ \sin \mleft(s+t\mright)=\sin \mleft(s\mright)\cos \mleft(t\mright)+\cos \mleft(s\mright)\sin \mleft(t\mright) \\ \sin ((\pi)/(2)+x)=\sin ((\pi)/(2))\cos (x)+\cos ((\pi)/(2))\sin (x) \\ we\text{ know that }\sin ((\pi)/(2))=1\text{ and }\cos ((\pi)/(2))=0.\text{ The,} \\ \sin ((\pi)/(2)+x)=1\cdot\cos (x)+0\cdot\sin (x) \\ \sin ((\pi)/(2)+x)=\cos (x) \\ \text{Therefore, let's replace the data} \\ sin((\pi)/(2)+x)\cdot\: tan(x)=\cos (x)\cdot\text{ tan}(x) \\ sin((\pi)/(2)+x)\cdot\: tan(x)=\cos (x)\cdot(\sin(x))/(\cos(x)) \\ sin((\pi)/(2)+x)\cdot\: tan(x)=\sin (x) \end{gathered}`

`\begin{gathered} b)\text{ }(\cos\left(-x\right))/(\sin\left(-x\right)) \\ It\text{ is important to know that} \\ \cos \mleft(-x\mright)=\cos \mleft(x\mright)\text{ and }\sin (-x)=\text{ -sin(x)} \\ \text{Therefore, let's replace the previous data} \\ (\cos(-x))/(\sin(-x))=(\cos(x))/(-\sin(x))=-(\cos(x))/(\sin(x))=-\cot (x) \end{gathered}`

`\begin{gathered} c)\text{ }\mleft(1-sin^2\mleft(x\mright)\mright)\cdot\: sec\mleft(x\mright) \\ It\text{ is important to know that} \\ \cos ^2(x)+sin^2(x)=1\to1-sin^2(x)=\cos ^2(x) \\ \text{Therefore, let's replace the previous data} \\ (1-sin^2(x))\cdot\: sec(x)=\cos ^2(x)\cdot\sec (x) \\ (1-sin^2(x))\cdot\: sec(x)=\cos ^2(x)\cdot\frac{1}{\cos\text{ (x)}} \\ (1-sin^2(x))\cdot\: sec(x)=\cos ^{}(x) \\ \end{gathered}`

`\begin{gathered} d)\sec \mleft(x\mright)-\sin \mleft(x\mright)\tan \mleft(x\mright) \\ \sec (x)-\sin (x)\tan (x)=(1)/(\cos(x))-\sin (x)\cdot(\sin(x))/(\cos(x)) \\ \sec (x)-\sin (x)\tan (x)=(1)/(\cos(x))-(\sin^2(x))/(\cos(x)) \\ \sec (x)-\sin (x)\tan (x)=(1-\sin^2(x))/(\cos(x)) \\ It\text{ is important to know that} \\ \cos ^2(x)+sin^2(x)=1\to1-sin^2(x)=\cos ^2(x) \\ Therefore, \\ \sec (x)-\sin (x)\tan (x)=(1-\sin^2(x))/(\cos(x)) \\ \sec (x)-\sin (x)\tan (x)=(\cos ^2(x))/(\cos (x))=\cos (x) \end{gathered}`