The derivatives of trigonometrical functions are given below:
Derivative of $\sin x$
Let $y=\sin x$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then \[y+\Delta y=\sin(x+\Delta x)\] \[\Delta y=\sin(x+\Delta x)-y\] \[\Delta y=\sin(x+\Delta x)-\sin x\] \[\Delta y=2\sin\frac{\Delta x}{2}.\cos\frac{2x+\Delta x}{2}\] \[\frac{\Delta y}{\Delta x}=\frac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}.\cos\frac{2x+\Delta x}{2}\] \[\therefore \frac{dy}{dx}=\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}\] \[=\lim_{\Delta x\to 0} \left(\frac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}.\cos\frac{2x+\Delta x}{2}\right)\] \[=\cos x\] \[\therefore \frac{d(\sin x)}{dx}=\cos x\]
Derivative of $\cos x$
Let $y=\cos x$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then \[y+\Delta y=\cos(x+\Delta x)\] \[\Delta y=\cos(x+\Delta x)-y\] \[\Delta y=\cos(x+\Delta x)-\cos x\] \[\Delta y=2\sin\frac{2x+\Delta x}{2}.\sin\left(-\frac{\Delta x}{2}\right)\] \[\Delta y=-2\sin\frac{2x+\Delta x}{2}.\sin\frac{\Delta x}{2}\] \[\frac{\Delta y}{\Delta x}=-\sin\frac{2x+\Delta x}{2}.\frac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}\] \[\therefore \frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}\] \[=\lim_{\Delta x\to 0}\left(-\sin\frac{2x+\Delta x}{2}.\frac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}\right)\] \[=-\sin x\] \[\therefore \frac{d(\cos x)}{dx}=-\sin x\]
Derivative of $\tan x$
Let $y=\tan x$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then \[y+\Delta y=\tan(x+\Delta x)\] \[\Delta y=\tan(x+\Delta x)-y\] \[\Delta y=\tan(x+\Delta x)-\tan x\] \[=\frac{\sin(x+\Delta x)}{\cos(x+\Delta x)}-\frac{\sin x}{\cos x}\] \[=\frac{\sin(x+\Delta x)\cos x-\cos(x+\Delta x)\sin x}{\cos(x+\Delta x)\cos x}\] \[=\frac{\sin(x+\Delta x-x)}{\cos(x+\Delta x)\cos x}\] \[=\frac{\sin\Delta x}{\cos(x+\Delta x)\cos x}\] \[\therefore \frac{\Delta y}{\Delta x}=\frac{\sin\Delta x}{\Delta x}.\frac{1}{\cos(x+\Delta x)\cos x}\] \[\therefore \frac{dy}{dx}=\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}\] \[=\lim_{\Delta x\to 0}\left(\frac{\sin\Delta x}{\Delta x}.\frac{1}{\cos(x+\Delta x)\cos x}\right)\] \[=\frac{1}{\cos x.\cos x}=\sec^2x\] \[\therefore \frac{d(\tan x)}{dx}=\sec^2x\]
Derivative of $\cot x$
Let $y=\cot x$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then \[y+\Delta y=\cot(x+\Delta x)\] \[\Delta y=\cot(x+\Delta x)-y\] \[\Delta y=\cot(x+\Delta x)-\cot x\] \[\Delta y=\frac{\cos (x+\Delta x)}{\sin (x+\Delta x)}-\frac{\cos x}{\sin x}\] \[=\frac{\sin x\cos(x+\Delta x)-\cos x\sin(x+\Delta x)}{\sin(x+\Delta x)\sin x}\] \[\Delta y=\frac{\sin(x-x-\Delta x)}{\sin(x+\Delta x)\sin x}\] \[\Delta y=\frac{-\sin\Delta x}{\sin(x+\Delta x)\sin x}\] \[\therefore \frac{\Delta y}{\Delta x}=-\frac{\sin\Delta x}{\Delta x}.\frac{1}{\sin(x+\Delta x)\sin x}\] \[\therefore \frac{dy}{dx}=\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}\] \[=\lim_{\Delta x\to 0}\left(-\frac{\sin\Delta x}{\Delta x}.\frac{1}{\sin(x+\Delta x)\sin x}\right)\] \[=-\frac{1}{\sin x\sin x}=- \operatorname{cosec}^2x\] \[\therefore \frac{d(\cot x)}{dx}=- \operatorname{cosec}^2x\]
Derivative of $\sec x$
Let $y=\sec x=\frac{1}{\cos x}$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then \[y+\Delta y=\frac{1}{\cos(x+\Delta x)}\] \[\Delta y=\frac{1}{\cos(x+\Delta x)-y}\] \[\Delta y=\frac{1}{\cos(x+\Delta x)}-\frac{1}{\cos x}\] \[=\frac{\cos x-\cos(x+\Delta x)}{\cos(x+\Delta x)\cos x}\] \[=\frac{2\sin\frac{2x+\Delta x}{2}\sin\frac{\Delta x}{2}}{\cos x\cos(x+\Delta x)}\] \[\therefore \frac{\Delta y}{\Delta x}=\frac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}.\frac{\sin\frac{2x+\Delta x}{2}}{\cos x\cos(x+\Delta x)}\] \[\therefore \frac{dy}{dx}=\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}\]\[=\lim_{\Delta x\to 0}\left\{\frac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}.\frac{\sin\frac{2x+\Delta x}{2}}{\cos x\cos(x+\Delta x)}\right\}\]\[=\frac{\sin x}{\cos x.\cos x}=\sec x.\tan x\]\[\therefore \frac{d(\sec x)}{dx}=\sec x.\tan x\]
Derivative of $\operatorname{cosec}x$
Let $y=\operatorname{cosec} x=\frac{1}{\sin x}$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then \[y+\Delta y=\frac{1}{\sin(x+\Delta x)}\] \[\Delta y=\frac{1}{\sin(x+\Delta x)}-y\]\[\Delta y=\frac{1}{\sin(x+\Delta x)}-\frac{1}{\sin x}\] \[\Delta y=\frac{\sin x-\sin(x+\Delta x)}{\sin(x+\Delta x)\sin x}\] \[\Delta y=\frac{2\sin\left(\frac{-\Delta x}{2}\right)\cos\frac{2x+\Delta x}{2}}{\sin(x+\Delta x)\sin x}\] \[\therefore \frac{\Delta y}{\Delta x}=\frac{-\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}.\frac{\cos\frac{2x+\Delta x}{2}}{\sin(x+\Delta x)\sin x}\] \[\therefore \frac{dy}{dx}=\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}\]\[=\lim_{\Delta x\to 0}\left\{-\frac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}.\frac{\cos\frac{2x+\Delta x}{2}}{\sin(x+\Delta x)\sin x}\right\}\] \[=-\frac{\cos x}{\sin x.\sin x}=-\operatorname{cosec}x.\cot x\]