Respuesta :

[tex]\begin{gathered} a)sin\mleft(\frac{\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 (\frac{\pi}{2}+x)=\sin (\frac{\pi}{2})\cos (x)+\cos (\frac{\pi}{2})\sin (x) \\ we\text{ know that }\sin (\frac{\pi}{2})=1\text{ and }\cos (\frac{\pi}{2})=0.\text{ The,} \\ \sin (\frac{\pi}{2}+x)=1\cdot\cos (x)+0\cdot\sin (x) \\ \sin (\frac{\pi}{2}+x)=\cos (x) \\ \text{Therefore, let's replace the data} \\ sin(\frac{\pi}{2}+x)\cdot\: tan(x)=\cos (x)\cdot\text{ tan}(x) \\ sin(\frac{\pi}{2}+x)\cdot\: tan(x)=\cos (x)\cdot\frac{\sin(x)}{\cos(x)} \\ sin(\frac{\pi}{2}+x)\cdot\: tan(x)=\sin (x) \end{gathered}[/tex]

[tex]\begin{gathered} b)\text{ }\frac{\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} \\ \frac{\cos(-x)}{\sin(-x)}=\frac{\cos(x)}{-\sin(x)}=-\frac{\cos(x)}{\sin(x)}=-\cot (x) \end{gathered}[/tex]

[tex]\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}[/tex]

[tex]\begin{gathered} d)\sec \mleft(x\mright)-\sin \mleft(x\mright)\tan \mleft(x\mright) \\ \sec (x)-\sin (x)\tan (x)=\frac{1}{\cos(x)}-\sin (x)\cdot\frac{\sin(x)}{\cos(x)} \\ \sec (x)-\sin (x)\tan (x)=\frac{1}{\cos(x)}-\frac{\sin^2(x)}{\cos(x)} \\ \sec (x)-\sin (x)\tan (x)=\frac{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)=\frac{1-\sin^2(x)}{\cos(x)} \\ \sec (x)-\sin (x)\tan (x)=\frac{\cos ^2(x)}{\cos (x)}=\cos (x) \end{gathered}[/tex]