Trigonometric Functions

$\therefore {\displaystyle \underset{-6}{\overset{0}{\int }}f\left(x\right)dx=2*\frac{1}{2}\left(2+5\right)*3=21}$

$16sin20°.sin40°.sin80°$

$=4sin60°\left\{?4sin\theta .sin\left(60°-\theta \right).sin\left(60°+\theta \right)=sin3\theta \right\}=2\sqrt[]{3}$

$co{s}^{-1}\frac{1}{\sqrt[]{5}}=co{t}^{-1}\frac{1}{2}$

$tan\left(2\left(ta{n}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{2}\right)\right)+ta{n}^{-1}\left(\frac{1}{2}\right)=tan\left(\pi +ta{n}^{-1}\frac{1}{2}\right)=\frac{1}{2}$

$tan2\alpha =2\sqrt[]{2}\Rightarrow \frac{2tan\alpha }{1-ta{n}^2\alpha }=2\sqrt[]{2}$

$\sqrt[]{2}ta{n}^2\alpha +tan\alpha -\sqrt[]{2}=0$

$tan\alpha =\frac{1}{\sqrt[]{2}},\left(-\sqrt[]{2}\right)\to rejected$

Draw y = cos²x and y = 2x + $\text{exception 25:}$

${\int }_0^3?g\left(x\right)-f\left(x\right)={\int }_0^3?\left|\right|x-2|-2|dx-{\int }_0^3?|x-2|dx$

$=\left(\frac{1}{2}*2*2+1+\frac{1}{2}*1*1\right)-\left(\frac{1}{2}*2*2+\frac{1}{2}*1*1\right)$

$=\left(2+1+\frac{1}{2}\right)-\left(2+\frac{1}{2}\right)=1$

 $\left(p\wedge q\right)\to \left(p\wedge r\right)$

$\equiv \sim \left(p\wedge q\right)\vee \left(p\wedge r\right)$

$\left(A\right)\left(\sim q\right)\vee \left(p\wedge r\right)\left(B\right)\left(\sim p\right)\vee \left(p\wedge r\right)$

$\left(C\right)\sim \left(p\wedge r\right)\vee \left(p\wedge q\right)\left(D\right)\sim \left(p\wedge q\right)\vee r$

Option (D) is correct

 ${\sum }_{r=0}^{20}?{}^{50-r}{\mathrm{C}}_6={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+{}^{30}{\mathrm{C}}_6$

$\begin{array}{rr}& ={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+\left({}^{30}{\mathrm{C}}_6+{}^{30}{\mathrm{C}}_7\right)-{}^{30}{\mathrm{C}}_7\\ & ={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+\left({}^{31}{\mathrm{C}}_6+{}^{31}{\mathrm{C}}_7\right)-{}^{30}{\mathrm{C}}_7\\ & ={}^{50}{\mathrm{C}}_6+{}^{50}{\mathrm{C}}_7-{}^{30}{\mathrm{C}}_7\\ & ={}^{51}{\mathrm{C}}_7-{}^{30}{\mathrm{C}}_7\end{array}$

$\left(\frac{\mathrm{K}-2}{\mathrm{h}-1}\right)\left(\frac{1-2}{3-1}\right)=-1\Rightarrow \mathrm{K}=2\mathrm{h}$

$?[?\mathrm{A}\mathrm{B}\mathrm{C}]=5\sqrt[]{5}$

$\Rightarrow \frac{1}{2}\left(\sqrt[]{5}\right)\sqrt[]{(\mathrm{h}-1{)}^2+(\mathrm{K}-2{)}^2}=5\sqrt[]{5}$

$\alpha =sin36°=x\left(say\right)$

$\therefore x=\frac{\sqrt[]{10-2\sqrt[]{5}}}{4}$

$\Rightarrow 16x^2=10-2\sqrt[]{5}$

$\Rightarrow 16x^4-80x^2+20=0$

$\therefore 4x^4-20x^2+5=0$

44. Secx = 2.                      .

We have, Secx = 2, | Secx is (+)ve  the principal solution lies in I^st and IV^thquadrent.

= $\frac{\pi }{3}$ and $\frac{6\pi -\pi }{3}$

= $\frac{\pi }{3}$ and $\frac{\mathrm{5\pi }}{3}$ .

As secx = sec $\frac{\pi }{3}$ . cosx = cos $\frac{\pi }{3}$ $\left[\therefore secx=\frac{1}{cosx}\right]$

The general solution is

x = 2nπ ±$\frac{\pi }{3}$, n ∈ z.