The value of cos
Question: The value of cos248 sin212 is ___________. Solution: $\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}$ $=\cos ^{2} 48^{\circ}-\cos ^{2}\left(90^{\circ}-12^{\circ}\right)$ $=\cos ^{2} 48^{\circ}-\cos ^{2} 78^{\circ}$ $=(\cos 48-\cos 78)(\cos 48+\cos 78)$ $=\left(2 \sin 63^{\circ} \sin 15^{\circ}\right)\left(2 \cos 63^{\circ} \cos 15^{\circ}\right)$ Using identities : $\cos A-\cos B=2 \sin \frac{(A+B)}{2} \sin \frac{(B-A)}{2}$ and $\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\fr...
Read More →In a triangle ABC with ∠C
Question: In a triangle $A B C$ with $\angle C=\frac{\pi}{2}$ the equation whose roots are $\tan A$ and $\tan B$ is. Solution: ${ }^{\mathrm{C}}$ Given in Triangel $A B C$, $\angle C=\pi / 2$ $\Rightarrow \angle A+\angle B=\pi / 2 \quad(\because \angle A+\angle B+\angle C=\pi$ sum of angle of triangle is $\pi)$ Equation with roots $\tan A$ and $\tan B$ is of the form, $(x-\tan A)(x-\tan B)=0$ i. e. $x^{2}=(\tan A+\tan B) x+\tan A \tan B=0$ Since $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$...
Read More →Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B
Question: Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units /kg of vitamin A and 5 units /kg of vitamin B while food Q contains 4 units /kg of vitamin A and 2 units /kg of vitamin B. Determine the minimum cost of the mixture? Solution: Let the mixture containxkg of food P andykg of food Q. Therefore, $x \geq ...
Read More →Maximise
Question: Maximise $Z=x+y$, subject to $x-y \leq-1,-x+y \leq 0, x, y \geq 0$. Solution: The region determined by the constraints, $x-y \leq-1,-x+y \leq 0, x, y \geq 0$, is as follows. There is no feasible region and thus, Z has no maximum value....
Read More →If k=sin
Question: If $k=\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}$, then the numerical value of $k$ is________________ Solution: Given, $\quad k=\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}$ $=\frac{1}{2}\left[2 \sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}\right]$ [using identity : $2 \sin A \sin B=\cos (A-B)-\cos (A+B)$ ] $=\frac{1}{2}\left\{\left[\cos \left(\frac{5 \pi}{18}-\frac{\pi}{18}\right)-\cos \left(\frac{5 \pi}{18}+\frac{\pi}{18}\righ...
Read More →Maximise Z = − x + 2y, subject to the constraints:
Question: Maximise $Z=-x+2 y$, subject to the constraints: $x \geq 3, x+y \geq 5, x+2 y \geq 6, y \geq 0$ Solution: The feasible region determined by the constraints, $x \geq 3, x+y \geq 5, x+2 y \geq 6$, and $y \geq 0$, is as follows. It can be seen that the feasible region is unbounded. The values of Z at corner points A (6, 0), B (4, 1), and C (3, 2) are as follows. As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value. For this, we graph the inequality, x+...
Read More →In a ∆ABC right angled at B, ∠A = ∠C. Find the values of
Question: In a $\triangle \mathrm{ABC}$ right angled at $\mathrm{B}, \angle \mathrm{A}=\angle \mathrm{C}$. Find the values of (i) sin A cos C + cos A sin C(ii) sin A sin B + cos A cos B Solution: (i) We have drawn the following figure related to given information To find: $\sin A \cos C+\cos A \sin C \ldots \ldots$ (1) Now we have, $\sin A=\frac{B C}{A C}, \sin A=\frac{A B}{A C}$ $\cos A=\frac{A B}{A C}, \cos C=\frac{B C}{A C}$ Now by substituting the above values in equation (1) We get, \sin A ...
Read More →Minimise and Maximise
Question: Minimise and Maximise $Z=x+2 y$ subject to $x+2 y \geq 100,2 x-y \leq 0,2 x+y \leq 200 ; x, y \geq 0$. Solution: The feasible region determined by the constraints,x+ 2y 100, 2xy 0, 2x+y 200,x 0, andy 0, is as follows. The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100), and D(0, 200). The values of Z at these corner points are as follows. The maximum value of Z is 400 at (0, 200) and the minimum value of Z is 100 at all the points on the line segment joining th...
Read More →If tan x
Question: If $\tan x=\frac{1-\cos y}{\sin y}$, then $\tan 2 x=$ Solution: Given, $\quad \tan x=\frac{1-\cos y}{\sin y}$ $\left[\right.$ using identity, $\left.\quad \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right]$ $\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}$ $=\frac{2\left(\frac{1-\cos y}{\sin y}\right)}{1-\left(\frac{1-\cos y}{\sin y}\right)^{2}}=\frac{2\left(\frac{2 \sin ^{2} y / 2}{2 \sin ^{y} / 2 \cos y / 2}\right)}{1-\left(\frac{2 \sin ^{2} y / 2}{2 \sin y / 2 \cos y / 2}\right)^...
Read More →Minimise and Maximise
Question: Minimise and Maximise $Z=5 x+10 y$ subject to $x+2 y \leq 120, x+y \geq 60, x-2 y \geq 0, x, y \geq 0$. Solution: The feasible region determined by the constraints,x+ 2y 120,x+y 60,x 2y 0,x 0, andy 0, is as follows. The corner points of the feasible region are A (60, 0), B (120, 0), C (60, 30), and D (40, 20). The values of Z at these corner points are as follows. The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on the line segment joining (1...
Read More →The value of cos
Question: The value of $\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}$ is ___________________ Solution: $\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}$ $=\cos 2\left(\frac{\pi}{15}\right) \cos 2^{2}\left(\frac{\pi}{15}\right) \cos 2^{3}\left(\frac{\pi}{15}\right) \cos \left(\pi+\frac{\pi}{15}\right)$ $[$ since, $\cos (\pi+\theta)=-\cos \theta]$ $=\cos 2\left(\frac{\pi}{15}\right) \cos 2^{2}\left(\frac{\pi}{15}\...
Read More →Minimise
Question: Minimise $Z=x+2 y$ subject to $2 x+y \geq 3, x+2 y \geq 6, x, y \geq 0$. Solution: The feasible region determined by the constraints, 2x+y 3,x+ 2y 6,x 0, andy 0, is as follows. The corner points of the feasible region are A (6, 0) and B (0, 3). The values of Z at these corner points are as follows. It can be seen that the value of Z at points A and B is same. If we take any other point such as (2, 2) on linex+ 2y= 6, then Z = 6 Thus, the minimum value of Z occurs for more than 2 points...
Read More →If sin (A−B)=12 and cos (A+B)=12, 0° <A+B≥90°, A<B find A and B.
Question: If $\sin (A-B)=\frac{1}{2}$ and $\cos (A+B)=\frac{1}{2}, 0^{\circ}A+B \geq 90^{\circ}, AB$ find $\mathrm{A}$ and $\mathrm{B}$. Solution: Given: $\sin (A-B)=\frac{1}{2} \ldots \ldots$(1) $\cos (A+B)=\frac{1}{2}$....(2) We know that, $\sin 30^{\circ}=\frac{1}{2}$....(3) $\cos 60^{\circ}=\frac{1}{2}$....(4) Now by comparing equation (1) and (3) We get, $A-B=30$......(5) Now by comparing equation (2) and (4) We get, $A+B=60 \ldots(6)$ Now to get the values ofAandB, let us solve equation (5...
Read More →Maximise
Question: Maximise $Z=3 x+2 y$ subject to $x+2 y \leq 10,3 x+y \leq 15, x, y \geq 0$. Solution: The feasible region determined by the constraints, $x+2 y \leq 10,3 x+y \leq 15, x \geq 0$, and $y \geq 0$, is as follows. The corner points of the feasible region are A (5, 0), B (4, 3), and C (0, 5). The values of Z at these corner points are as follows. Therefore, the maximum value of Z is 18 at the point (4, 3)....
Read More →Maximise
Question: Maximise $Z=3 x+2 y$ subject to $x+2 y \leq 10,3 x+y \leq 15, x, y \geq 0$. Solution: The feasible region determined by the constraints, $x+2 y \leq 10,3 x+y \leq 15, x \geq 0$, and $y \geq 0$, is as follows. The corner points of the feasible region are A (5, 0), B (4, 3), and C (0, 5). The values of Z at these corner points are as follows. Therefore, the maximum value of Z is 18 at the point (4, 3)....
Read More →If cos x cos 2x cos2
Question: If $\cos x \cos 2 x \cos 2^{2} x_{\ldots} \cos 2^{n-1} x=\lambda \frac{\sin 2^{n} \lambda}{\sin x}$, then $\lambda=$ ______________________ Solution: $\cos x \cos 2 x \cos 2^{2} x \ldots \ldots \ldots \ldots \cos 2_{x}^{n-1}$ multiply and divide the above equation by $2 \sin x$ we get, $=\frac{1}{2 \sin x}\left[2 \sin x \cos x \cos 2 x \cos 2^{2} x \ldots \ldots \ldots \cos 2^{n-1} x\right]$ (using identity, $2 \sin x \cos x=\sin 2 x$ ) $=\frac{1}{2 \sin x}\left[\sin 2 x \cos 2 x \cos ...
Read More →Minimise
Question: Minimise $Z=3 x+5 y$ such that $x+3 y \geq 3, x+y \geq 2, x, y \geq 0$. Solution: The feasible region determined by the system of constraints, $x+3 y \geq 3, x+y \geq 2$, and $x, y \geq 0$, is as follows. It can be seen that the feasible region is unbounded. The corner points of the feasible region are $A(3,0), B\left(\frac{3}{2}, \frac{1}{2}\right)$, and $C(0,2)$. The values of Z at these corner points are as follows. As the feasible region is unbounded, therefore, 7 may or may not be...
Read More →If tan (A−B)=13√ and tan (A+B)=3–√, 0° < A+B≤90°, A>B find A and B.
Question: If $\tan (A-B)=\frac{1}{\sqrt{3}}$ and $\tan (A+B)=\sqrt{3}, 0^{\circ}A+B \leq 90^{\circ}, AB$ find $\mathrm{A}$ and $\mathrm{B}$. Solution: Given: $\tan (A-B)=\frac{1}{\sqrt{3}}$......(1) $\tan (A+B)=\sqrt{3}$......(2) We know that, $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$.....(3) $\tan 60^{\circ}=\sqrt{3}$......(4) Now by comparing equation (1) and (3) We get, $A-B=30$....(5) Now by comparing equation (2) and (4) We get, $A+B=60 \ldots \ldots(6)$ Now to get the values ofAandB, let us sol...
Read More →If
Question: If $\frac{1-\tan ^{2}\left(\frac{\pi}{4}-x\right)}{1+\tan ^{2}\left(\frac{\pi}{4}-x\right)}=\sin k x$, then $k=$ ____________________ Solution: Given, $\frac{1-\tan ^{2}(\pi / 4-x)}{1+\tan ^{2}(\pi / 4-x)}=\sin k x$ $\left[\right.$ using identity $\left.: \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta\right]$ We have, $\frac{1-\tan ^{2}(\pi / 4-x)}{1+\tan ^{2}(\pi / 4-x)}=\cos 2(\pi / 4-x)$ $=\cos (\pi / 2-2 x)$ $=\sin 2 x(\because \cos (\pi / 2-\theta)=\sin \theta)$ $=\si...
Read More →Maximise
Question: Maximise Z = 5x+ 3y subject to $3 x+5 y \leq 15,5 x+2 y \leq 10, x \geq 0, y \geq 0$. Solution: The feasible region determined by the system of constraints, $3 x+5 y \leq 15$, $5 x+2 y \leq 10, x \geq 0$, and $y \geq 0$, are as follows. The corner points of the feasible region are $\mathrm{O}(0,0), \mathrm{A}(2,0), \mathrm{B}(0,3)$, and $\mathrm{C}\left(\frac{20}{19}, \frac{45}{19}\right)$. The values of Z at these corner points are as follows. Therefore, the maximum value of $Z$ is $\...
Read More →The value of cos 12° + cos 84° + cos 156° + cos 132° is
Question: The value of cos 12 + cos 84 + cos 156 + cos 132 is (a) $\frac{1}{2}$ (b) 1 (C) $-\frac{1}{2}$ (d) $-\frac{1}{8}$ Solution: $\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}$ $=\cos 12^{\circ}+\cos 84^{\circ}+\cos \left(180^{\circ}-24\right)+\cos \left(108^{\circ}-4 \mathrm{~F}\right)$ $=\cos 12^{\circ}+\cos 84^{\circ}-\cos 24^{\circ}-\cos 48^{\circ}(\because \cos (180-\theta)=-\cos \theta)$ $=\cos 12^{\circ}-\cos 48^{\circ}+\cos 84^{\circ}-\cos 24^{\circ}$ $=-2 \sin \...
Read More →The value of cos 12° + cos 84° + cos 156° + cos 132° is
Question: The value of cos 12 + cos 84 + cos 156 + cos 132 is (a) $\frac{1}{2}$ (b) 1 (C) $-\frac{1}{2}$ (d) $-\frac{1}{8}$ Solution: $\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}$ $=\cos 12^{\circ}+\cos 84^{\circ}+\cos \left(180^{\circ}-24\right)+\cos \left(108^{\circ}-4 \mathrm{~F}\right)$ $=\cos 12^{\circ}+\cos 84^{\circ}-\cos 24^{\circ}-\cos 48^{\circ}(\because \cos (180-\theta)=-\cos \theta)$ $=\cos 12^{\circ}-\cos 48^{\circ}+\cos 84^{\circ}-\cos 24^{\circ}$ $=-2 \sin \...
Read More →Minimise
Question: Minimise Z = 3x+ 4y subject to $x+2 y \leq 8,3 x+2 y \leq 12, x \geq 0, y \geq 0$. Solution: The feasible region determined by the system of constraints, $x+2 y \leq 8,3 x+2 y \leq 12, x \geq 0$, and $y \geq 0$, is as follows. The corner points of the feasible region are O (0, 0), A (4, 0), B (2, 3), and C (0, 4). The values of Z at these corner points are as follows. Therefore, the minimum value of Z is 12 at the point (4, 0)....
Read More →If sin (A + B) = 1 and cos (A − B) = 1,
Question: If $\sin (A+B)=1$ and $\cos (A-B)=1,0^{\circ}A+B \leq 90^{\circ}, A \geq B$ find $A$ and $B$. Solution: Given: $\sin (A+B)=1$...(1) $\cos (A-B)=1$....(2) We know that, $\sin 90^{\circ}=1$.....(3) $\cos 0^{\circ}=1$.....(4) Now by comparing equation (1) and (3) We get, $A+B=90 \ldots \ldots(5)$ Now by comparing equation (2) and (4) We get, $A-B=0 \ldots \ldots(6)$ Now to get the values ofAandB, let us solve equation (5) and (6) simultaneously Therefore by adding equation (5) and (6) We ...
Read More →If sin θ=
Question: If $\sin \theta=-\frac{4}{5}$ and $\theta$ lies in third quadrant, then the value of $\cos \frac{\theta}{2}$ is (a) $\frac{1}{5}$ (b) $-\frac{1}{\sqrt{10}}$ (c) $-\frac{1}{\sqrt{5}}$ (d) $\frac{1}{\sqrt{10}}$ Solution: $\sin \theta=\frac{-4}{5}$ where $\theta$ lies in third quadrant. since $\cos ^{2} \theta=1-\sin ^{2} \theta$ $=1-\left(\frac{-4}{5}\right)^{2}$ $=1-\frac{16}{25}$ $\cos ^{2} \theta=\frac{9}{25}$ i. e. $\cos \theta=\pm \frac{3}{5}$ since $\pi\theta\frac{3 \pi}{2}$ i. e. ...
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