Prove that: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
[question] Question. Prove that: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$ [/question] [solution] solution: L.H.S. $=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$ $=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \left(\frac{\frac{3 \pi}{13}+\frac{5 \pi}{13}}{2}\right) \cos \left(\frac{\frac{3 \pi}{13}-\frac{5 \pi}{13}}{2}\right)\left[\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-...
Read More →Prove that $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
[question] Question. Prove that $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$ [/question] [solution] solution: It is known that $\tan 2 \mathrm{~A}=\frac{2 \tan \mathrm{A}}{1-\tan ^{2} \mathrm{~A}}$. $\therefore$ L.H.S. $=\tan 4 x=\tan 2(2 x)$ $=\frac{2 \tan 2 x}{1-\tan ^{2}(2 x)}$ $=\frac{2\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)}{1-\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)^{2}}$ $=\frac{\left(\frac{4 \tan x}{1-\tan ^{2} x}\right)}{\left[1-\frac{4 \tan ^...
Read More →Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$
[question] Question. Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$ [/question] [solution] solution: L.H.S. $=\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}$ $=\frac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x}$ $=\frac{2 \cos \left(\frac{4 \mathrm{x}+2 \mathrm{x}}{2}\right) \cos \left(\frac{4 \mathrm{x}-2 \mathrm{x}}{2}\right)+\cos 3 \mathrm{x}}{2 \sin \left(\frac{4 \mathrm{x}+2 \mathrm{x}}{2}\right) \cos \left(\frac{4 \mathrm{x...
Read More →Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
[question] Question. Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$ [/question] [solution] solution: It is known that $\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos ^{2} A-\sin ^{2} A=\cos 2 A$ $\therefore$ L.H.S. $=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$ $=\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}$ $=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}$ $=-2 \times(-\sin x)$ $=2 \sin x=$ R.H....
Read More →Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
[question] Question. Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$ [/question] [solution] solution: It is known that $\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $\therefore$ L.H.S. $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$ $=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}...
Read More →Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$
[question] Question. Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$ [/question] [solution] solution: It is known that $\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $\therefore$ L.H.S. $=\frac{\sin x-\sin y}{\cos x+\cos y}$ $=\frac{2 \cos \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x...
Read More →Prove that $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
[question] Question. Prove that $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$ [/question] [solution] solution: It is known that $\sin \mathrm{A}+\sin \mathrm{B}=2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right), \cos \mathrm{A}+\cos \mathrm{B}=2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$ $\therefore$ L.H.S. $=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$ $=\frac{2 \sin \left(\fra...
Read More →Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
[question] Question. Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$ [/question] [solution] solution: It is known that $\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$ $\therefore$ L.H.S $=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}$ $=\frac{-2 \sin \left(\frac{9 x+5 x}{2}\right) \cdot \sin \left(\frac{9 x-5 x}{2}\right)}{2 \cos \left(\frac{17 ...
Read More →Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$
[question] Question. Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$ [/question] [solution] solution: It is known that $\cos \mathrm{A}-\cos \mathrm{B}=-2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$. $\therefore$ L.H.S. $=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$ $=-2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}...
Read More →$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
[question] Question. $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$ [/question] [solution] solution: L.H.S. $=\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$ $=\sin x \cos x[\tan x+\cot x]$ $=\sin x \cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$ $=(\sin x \cos x)\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right]$ $=1=\mathrm{R} . \math...
Read More →Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$
[question] Question. Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$ [/question] [solution] solution: L.H.S. $=\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}$ $=\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}$ $=\frac{-\cos ^{2} x}{-\sin ^{2} x}$ $=\cot ^{2} x$ $=$ R.HS. [/solution]...
Read More →Prove that: $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
[question] Question. Prove that: $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$ [/question] [solution] solution: $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$ $=\frac{1}{2}\left[2 \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\right]+\frac{1}{2}\left[-2 \sin \left(\frac{\pi}{4}-x\righ...
Read More →Prove that $2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}=10$
[question] Question. Prove that $2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}=10$ [/question] [solution] solution: L.H.S $=2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}$ $=2\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}^{2}+2\left(\frac{1}{\sqrt{2}}\right)^{2}+2(2)^{2}$ $=2\left\{\sin \frac{\pi}{4}\right\}^{2}+2 \times \frac{1}{2}+8$ $=2\left(\frac{1}{\sqrt{2}}\right)^{2}+1+8$ $=1+1+8$ $=10$ $=$ R.H.S [/solution]...
Read More →Prove that $\cot ^{2} \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}=6$
[question] Question. Prove that $\cot ^{2} \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}=6$ [/question] [solution] solution: L.H.S. $=\cot ^{2} \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}$ $=(\sqrt{3})^{2}+\operatorname{cosec}\left(\pi-\frac{\pi}{6}\right)+3\left(\frac{1}{\sqrt{3}}\right)^{2}$ $=3+\operatorname{cosec} \frac{\pi}{6}+3 \times \frac{1}{3}$ $=3+2+1=6$ $=$ R.H.S [/solution]...
Read More →Prove that $2 \sin ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$
[question] Question. Prove that $2 \sin ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$ [/question] [solution] solution: L.H.S. $=2 \sin ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}$ $=2\left(\frac{1}{2}\right)^{2}+\operatorname{cosec}^{2}\left(\pi+\frac{\pi}{6}\right)\left(\frac{1}{2}\right)^{2}$ $=2 \times \frac{1}{4}+\left(-\operatorname{cosec} \frac{\pi}{6}\right)^{2}\left(\frac{1}{4}\right)$ $=\frac{1}{2...
Read More →$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
[question] Question. $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$ [/question] [solution] solution: L.H.S. $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$ $=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$ $=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$ $=$ R.H.S. [/solution]...
Read More →Assuming complete dissociation, calculate the pH of the following solutions:
Question: Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH Solution: (i)0.003MHCl: $\mathrm{H}_{2} \mathrm{O}+\mathrm{HCl} \longleftrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}$ Since HCl is completely ionized, $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=[\mathrm{HCl}] .$ $\Rightarrow\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.003$ Now, $\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\...
Read More →Show that the function f:
Question: Show that the function $f . \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^{3}$ is injective. Solution: $f: \mathbf{R} \rightarrow \mathbf{R}$ is given as $f(x)=x^{3}$. Suppose $f(x)=f(y)$, where $x, y \in \mathbf{R}$. $\Rightarrow x^{3}=y^{3} \ldots(1)$ Now, we need to show thatx=y. Supposexy, their cubes will also not be equal. $\Rightarrow x^{3} \neq y^{3}$ However, this will be a contradiction to (1). x=y Hence,fis injective....
Read More →Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) =,
Question: Show that function $f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1x1\}$ defined by $f(x)=\frac{x}{1+|x|}, x \in \mathbf{R}$ is one-one and onto function. Solution: It is given that $f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1x1\}$ is defined as $f(x)=\frac{x}{1+|x|}, x \in \mathbf{R}$. Suppose $f(x)=f(y)$, where $x, y \in \mathbf{R}$. $\Rightarrow \frac{x}{1+|x|}=\frac{y}{1+|y|}$ It can be observed that ifxis positive andyis negative, then we have: $\frac{x}{1+x}=\frac{y}{1-y} \Righ...
Read More →The incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves Rs 1250, find their incomes.
Question: The incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves Rs 1250, find their incomes. Solution: Let the income ofbe Rsand the income ofbe Rs.further let the expenditure ofbeand the expenditure ofberespectively then, Saving of $x=8 x-19 y$ Saving of $Y=7 x-16 y$ $8 x-19 y=1250$ $7 x-16 y=1250$ $8 x-19 y-1250=0 \cdots(i)$ $7 x-16 y-1250=0 \cdots(i i)$ Solving equationandby cross- multiplication, we have $\frac{x}{(-19 \times-1250)-(-...
Read More →In a rectangle, if the length is increased by 3 meters and breadth is decreased by 4 meters, the area of the rectangle is reduced by 67 square meters
Question: In a rectangle, if the length is increased by 3 meters and breadth is decreased by 4 meters, the area of the rectangle is reduced by 67 square meters. If length is reduced by 1 meter and breadth is increased by 4 meters, the area is increased by 89 Sq. meters. Find the dimensions of the rectangle. Solution: Let the length and breadth of the rectangle beandunits respectively Then, area of rectangle =square units If the length is increased bymeters and breath is reduced each bysquare met...
Read More →What is the net flux of the uniform electric field
Question: What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? Solution: All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero....
Read More →Consider a uniform electric field
Question: Consider a uniform electric field $\mathbf{E}=3 \times 10^{3} \hat{\mathrm{N}} / \mathrm{C}$. (a) What is the flux of this field through a square of $10 \mathrm{~cm}$ on a side whose plane is parallel to the $y z$ plane? (b) What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the $x$-axis? Solution: (a) Electric field intensity, $\vec{E}=3 \times 10^{3} \hat{\mathrm{N}} \mathrm{N} / \mathrm{C}$ Magnitude of electric field intensity, $|\ve...
Read More →If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x)).
Question: If $f \mathbf{R} \rightarrow \mathbf{R}$ is defined bv $f(x)=x^{2}-3 x+2$ find $f(f(x))$ Solution: It is given that $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x^{2}-3 x+2$ $f(f(x))=f\left(x^{2}-3 x+2\right)$ $=\left(x^{2}-3 x+2\right)^{2}-3\left(x^{2}-3 x+2\right)+2$ $=x^{4}+9 x^{2}+4-6 x^{3}-12 x+4 x^{2}-3 x^{2}+9 x-6+2$ $=x^{4}-6 x^{3}+10 x^{2}-3 x$...
Read More →Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field.
Question: Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio? Solution: Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 mov...
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