In triangle ABC, prove the following:

Question: In triangle ABC, prove the following: $a \cos A+b \cos B+c \cos C=2 b \sin A \sin C=2 c \sin A \sin B$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \quad \ldots$ (1) Consider the LHS of the equation $a \cos A+b \cos B+c \cos C$. $a \cos A+b \cos B+c \cos C=k(\sin A \cos A+\sin B \cos B+\sin C \cos C)$ $=\frac{k}{2}(2 \sin A \cos A+2 \sin A \cos A+2 \sin C \cos C)$ $=\frac{k}{2}(\sin 2 A+\sin 2 B+\sin 2 C)$ $=\frac{k}{2}[2 \sin (A+B) \cos (A-B)+2 \sin C \cos C]$ $...

Given that E and F are events such that

Question: Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E F) = 0.2, find P (E|F) and P(F|E). Solution: It is given that P(E) = 0.6, P(F) = 0.3, and P(E F) = 0.2 $\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{0.2}{0.3}=\frac{2}{3}$ $\Rightarrow \mathrm{P}(\mathrm{F} \mid \mathrm{E})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{E})}=\frac{0.2}{0.6}=\frac{1}{3}$...

Find three different irrational numbers between the rational numbers

Question: Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$. Solution: As, $\frac{5}{7} \approx 0.714$ and $\frac{9}{11} \approx 0.818$ So, the three different irrational numbers are: 0.72020020002..., 0.7515511555111... and 0.808008000...Disclaimer:There are an infinite number of irrational numbers between two rational numbers....

In triangle ABC, prove the following:

Question: In triangle ABC, prove the following: $a \cos A+b \cos B+c \cos C=2 b \sin A \sin C=2 c \sin A \sin B$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \quad \ldots$ (1) Consider the LHS of the equation $a \cos A+b \cos B+c \cos C$. $a \cos A+b \cos B+c \cos C=k(\sin A \cos A+\sin B \cos B+\sin C \cos C)$ $=\frac{k}{2}(2 \sin A \cos A+2 \sin A \cos A+2 \sin C \cos C)$ $=\frac{k}{2}(\sin 2 A+\sin 2 B+\sin 2 C)$ $=\frac{k}{2}[2 \sin (A+B) \cos (A-B)+2 \sin C \cos C]$ $...

Find two rational and two irrational number between 0.5 and 0.55.

Question: Find two rational and two irrational number between 0.5 and 0.55. Solution: The two rational numbers between 0.5 and 0.55 are: 0.51 and 0.52The two irrational numbers between 0.5 and 0.55 are: 0.505005000... and 0.5101100111000...Disclaimer:There are infinite number of rational and irrational numbers between 0.5 and 0.55....

How many irrational numbers lie between

Question: How many irrational numbers lie between $\sqrt{2}$ and $\sqrt{3}$ ? Find any three irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}$. Solution: There are infinite number of irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}$. As, $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$ So, the three irrational numbers lying between $\sqrt{2}$ and $\sqrt{3}$ are: 1.420420042000..., 1.505005000... and 1.616116111......

If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =

Question: If $x \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)=\cos \left(90^{\circ}-\theta\right)$, then $x=$ (a) 0 (b) 1 (c) $-1$ (d) 2 Solution: We have: $x \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)=\cos \left(90^{\circ}-\theta\right)$ Here we have to find the value of $x$ We know that $\left[\begin{array}{l}\sin \left(90^{\circ}-\theta\right)=\cos \theta \\ \cos \left(90^{\circ}-\theta\right)=\sin \theta \\ \cot \left(90^{\circ}-\theta\r...

If A and B are complementary angles, then

Question: If A and B are complementary angles, then (a) sin A = sin B(b) cos A = cos B(c) tan A = tan B(d) sec A = cosec B Solution: Given: $A$ and are $B$ are complementary angles Since $\sec \left(90^{\circ}-B\right)=\operatorname{cosec} B$ therefore $A+B=90^{\circ}$ $\Rightarrow A=90^{\circ}-B$ $\Rightarrow \sec A=\sec \left(90^{\circ}-B\right)$ $\Rightarrow \sec A=\operatorname{cosec} B$ Hence the correct option is $(d)$...

Insert a rational and an irrational number between 2 and 2.5.

Question: Insert a rational and an irrational number between 2 and 2.5. Solution: As, few rational numbers between 2 and 2.5 are: 2.1, 2.2, 2.3, 2.4, ... And, Since, $2=\sqrt{4}$ and $2.5=\sqrt{6.25}$ So, irrational number between 2 ans $2.5$ are: $\sqrt{4.1}, \sqrt{4.2}, \ldots, \sqrt{5}, \ldots$ Hence, a rational and an irrational number can be $2.1$ and $\sqrt{5}$, respectively. Disclaimer: There are infinite rational and irrational numbers between any two rational numbers....

A toy company manufactures two types of dolls, A and B.

Question: A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many o...

If x cosec2 30° sec2 45°8 cos2 45° sin2 60°=tan2 60°−tan2 30°, then x =

Question: If $\frac{x \operatorname{cosec}^{2} 30^{\circ} \sec ^{2} 45^{\circ}}{8 \cos ^{2} 45^{\circ} \sin ^{2} 60^{\circ}}=\tan ^{2} 60^{\circ}-\tan ^{2} 30^{\circ}$, then $x=$ (a) 1 (b) $-1$ (c) 2 (d) 0 Solution: We have: $\frac{x \operatorname{cosec}^{2} 30^{\circ} \sec ^{2} 45^{\circ}}{8 \cos ^{2} 45^{\circ} \sin ^{2} 60^{\circ}}=\tan ^{2} 60^{\circ}-\tan ^{2} 30^{\circ}$ Here we have to find the value of $x$ As we know that $\left[\begin{array}{l}\cos 45^{\circ}=\frac{1}{\sqrt{2}} \\ \sec ...

In ∆ABC, prove that:

Question: In ∆ABC, prove that: $\frac{b \sec B+c \sec C}{\tan B+\tan C}=\frac{c \sec C+a \sec A}{\tan C+\tan A}=\frac{a \sec A+b \sec B}{\tan A+\tan B}$ Solution: Let ABC be any triangle. Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ Now, $\frac{b \sec B+c \sec C}{\tan B+\tan C}=\frac{\frac{b}{\cos B}+\frac{c}{\cos C}}{\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}}$ $=\frac{b \cos C+c \cos B}{\sin B \cos C+\sin C \cos B}$ $=\frac{k \sin B \cos C+k \sin C \cos B}{\sin B \cos C+\sin...

In ∆ABC, prove that:

Question: In ∆ABC, prove that: $\frac{b \sec B+c \sec C}{\tan B+\tan C}=\frac{c \sec C+a \sec A}{\tan C+\tan A}=\frac{a \sec A+b \sec B}{\tan A+\tan B}$ Solution: Let ABC be any triangle. Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ Now, $\frac{b \sec B+c \sec C}{\tan B+\tan C}=\frac{\frac{b}{\cos B}+\frac{c}{\cos C}}{\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}}$ $=\frac{b \cos C+c \cos B}{\sin B \cos C+\sin C \cos B}$ $=\frac{k \sin B \cos C+k \sin C \cos B}{\sin B \cos C+\sin...

The value of cos3 20°−cos3 70°sin3 70°−sin3 20°is

Question: The value of $\frac{\cos ^{3} 20^{\circ}-\cos ^{3} 70^{\circ}}{\sin ^{3} 70^{\circ}-\sin ^{3} 20^{\circ}}$ is (a) $\frac{1}{2}$ (b) $\frac{1}{\sqrt{2}}$ (c) 1 (d) 2 Solution: We have to evaluate the value. The formula to be used, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$ $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$ So, $=\frac{\cos ^{3} 20^{\circ}-\cos ^{3} 70^{\circ}}{\sin ^{3} 70^{\circ}-\sin ^{3} 20^{\circ}}$ $=\frac{\left(\cos 20^{\circ}-\cos 70\right)\left(\cos ^{2} 20^{\...

Refer to question 8.

Question: Refer to question 8. If the grower wants to maximize the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added? Solution: Let the fruit grower usexbags of brand P andybags of brand Q. The problem can be formulated as follows. Maximize $z=3 x+3.5 y$ (1) subject to the constraints, $x+2 y \geq 240$ (2) $x+0.5 y \geq 90$ (3) $1.5 x+2 y \leq 310$ (4) $x, y \geq 0$ (5) The feasible region determined by the system of...

The value of cos2 17° − sin2 73° is

Question: The value of $\cos ^{2} 17^{\circ}-\sin ^{2} 73^{\circ}$ is (a) 1 (b) $\frac{1}{3}$ (c) 0 (d) $-1$ Solution: We have: $\cos ^{2} 17^{\circ}-\sin ^{2} 73^{\circ}$ $=\cos ^{2}\left(90^{\circ}-73^{\circ}\right)-\sin ^{2} 73^{\circ}$ $=\sin ^{2} 73^{\circ}-\sin ^{2} 73^{\circ}$ $=0$ Hence the correct option is $(c)$...

If tan2 45° − cos2 30° = x sin 45° cos 45°, then x =

Question: If $\tan ^{2} 45^{\circ}-\cos ^{2} 30^{\circ}=x \sin 45^{\circ} \cos 45^{\circ}$, then $x=$ (a) 2 (b) $-2$ (c) $-\frac{1}{2}$ (d) $\frac{1}{2}$ Solution: We are given: $\tan ^{2} 45^{\circ}-\cos ^{2} 30^{\circ}=x \sin 45^{\circ} \cos 45^{\circ}$ We have to find $x$ $\Rightarrow 1-\left(\frac{\sqrt{3}}{2}\right)^{2}=x \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$ $\Rightarrow 1-\frac{3}{4}=\frac{x}{2}$ $\Rightarrow \frac{1}{4}=\frac{x}{2}$ $\Rightarrow x=\frac{1}{2}$ We know that $\left...

Examine whether the following numbers are rational or irrational.

Question: Examine whether the following numbers are rational or irrational. (i) $3+\sqrt{3}$ (ii) $\sqrt{7}-2$ (iii) $\sqrt[3]{5} \times \sqrt[3]{25}$ (iv) $\sqrt{7} \times \sqrt{343}$ (v) $\sqrt{\frac{13}{117}}$ (vi) $\sqrt{8} \times \sqrt{2}$ Solution: (i) Let us assume, to the contrary that $3+\sqrt{3}$ is rational. Then, $3+\sqrt{3}=\frac{p}{q}$, where $p$ and $q$ are coprime and $q \neq 0$. $\Rightarrow \sqrt{3}=\frac{p}{q}-3$ $\Rightarrow \sqrt{3}=\frac{p-3 q}{q}$ Since,pandqare are intege...

Question: In ∆ABC, prove that: $a \sin \frac{A}{2} \sin \left(\frac{B-C}{2}\right)+b \sin \frac{B}{2} \sin \left(\frac{C-A}{2}\right)+c \sin \frac{C}{2} \sin \left(\frac{A-B}{2}\right)=0$ Solution: Consider $a \sin \frac{A}{2} \sin \left(\frac{B-C}{2}\right)+b \sin \frac{B}{2} \sin \left(\frac{C-A}{2}\right)+c \sin \frac{C}{2} \sin \left(\frac{A-B}{2}\right)$ $=k\left[\sin A \sin \frac{A}{2} \sin \left(\frac{B-C}{2}\right)+\sin B \sin \frac{B}{2} \sin \left(\frac{C-A}{2}\right)+\sin C \sin \frac...

If 3 cos θ = 5 sin θ, then the value of 5 sin θ−2 sec3 θ+2 cos θ5 sin θ+2 sec3 θ−2 cos θis

Question: If $3 \cos \theta=5 \sin \theta$, then the value of $\frac{5 \sin \theta-2 \sec ^{3} \theta+2 \cos \theta}{5 \sin \theta+2 \sec ^{3} \theta-2 \cos \theta}$ is (a) $\frac{271}{979}$ (b) $\frac{316}{2937}$ (c) $\frac{542}{2937}$ (d) None of these Solution: We have, $3 \cos \theta=5 \sin \theta$ So we can manipulate it as, $\tan \theta=\frac{3}{5}$ So now we can get the values of other trigonometric ratios, $\sin \theta=\frac{3}{\sqrt{34}}$ $\cos \theta=\frac{5}{\sqrt{34}}$ $\sec \theta=\...

A fruit grower can use two types of fertilizer in his garden, brand P and brand Q.

Question: A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid at least 270 kg of potash and at most 310 kg of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen ...

In triangle ABC, prove the following:

Question: In triangle ABC, prove the following: $\frac{\cos ^{2} B-\cos ^{2} C}{b+c}+\frac{\cos ^{2} C-\cos ^{2} A}{c+a}+\frac{\cos ^{2} A-\cos ^{2} B}{a+b}=0$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ Then, Consider the LHS of the equation $\frac{\cos ^{2} B-\cos ^{2} C}{b+c}+\frac{\cos ^{2} C-\cos ^{2} A}{c+a}+\frac{\cos ^{2} A-\cos ^{2} B}{a+b}=0$. $\mathrm{LHS}=\frac{\cos ^{2} B-\cos ^{2} C}{b+c}+\frac{\cos ^{2} C-\cos ^{2} A}{c+a}+\frac{\cos ^{2} A-\cos ^{2} B}{a+...

If θ is an acute angle such that tan2 θ=87, then the value of (1+sin θ) (1−sin θ )(1+cos θ) (1−cos θ)is

Question: If $\theta$ is an acute angle such that $\tan ^{2} \theta=\frac{8}{7}$, then the value of $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ is (a) $\frac{7}{8}$ (b) $\frac{8}{7}$ (c) $\frac{7}{4}$ (d) $\frac{64}{49}$ Solution: Given that: $\tan ^{2} \theta=\frac{8}{7}$ and $\theta$ is an acute angle We have to find the following expression $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ Since $\tan ^{2} \theta=\frac{8}{7}$ $\tan \theta=\sqr...

In triangle ABC, prove the following:

Question: In triangle ABC, prove the following: $\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}-\frac{1}{a^{2}}-\frac{1}{b^{2}}$ Solution: Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ Then, Consider the LHS of the equation $\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}-\frac{1}{a^{2}}-\frac{1}{b^{2}}$. $\mathrm{LHS}=\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}$ $=\frac{1-2 \sin ^{2} A}{a^{2}}-\frac{1-2 \sin ^{2} B}{b^{2}}$ $=\frac{1-2 \frac{a^{2}}{k^{2}}}{a^{2}}-\frac{1-2 \frac{b^{...

If tan θ=34, then cos2 θ − sin2 θ =

Question: If $\tan \theta=\frac{3}{4}$, then $\cos ^{2} \theta-\sin ^{2} \theta=$ (a) $\frac{7}{25}$ (b) 1 (C) $\frac{-7}{25}$ (d) $\frac{4}{25}$ Solution: Given that: $\tan \theta=\frac{3}{4}$ Since $\tan x=\frac{\text { Perpendicular }}{\text { Base }}$ $\Rightarrow$ Perpendicular $=3$ $\Rightarrow$ Base $=4$ $\Rightarrow$ Hypotenuse $=\sqrt{9+16}$ $\Rightarrow$ Hypotenuse $=5$ We know that $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ and $\cos \theta=\frac{\text { Base }...