In a ∆ABC, if
Question: In a $\Delta A B C$, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=2$, then area of $\triangle A B C$ is equal to ___________________ Solution: In a $\Delta A B C$ Given $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c} \ldots(1)$ and $a=2$ Using sine formula $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \quad \ldots(2)$ using above two, (1) and (2) we get, $\frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}=\frac{\cos C}{\sin C}$ $\Rightarrow \cot A=\cot B=\cot C$ $\...
Read More →In a ∆ABC, if
Question: In a $\Delta A B C$, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=2$, then area of $\triangle A B C$ is equal to Solution: In a $\Delta A B C$ Given $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c} \ldots(1)$ and $a=2$ Using sine formula $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \quad \ldots(2)$ using above two, (1) and (2) we get, $\frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}=\frac{\cos C}{\sin C}$ $\Rightarrow \cot A=\cot B=\cot C$ $\Rightarrow$ angle $A...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=1$ Solution: We have to prove $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=1$ We know that, $\sin ^{2} A+\cos ^{2} A=1$ So, $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=\sin ^{2} A \frac{\cos ^{2} A}{\sin ^{2} A}+\cos ^{2} A \frac{\sin ^{2} A}{\cos ^{2} A}$ $=\cos ^{2} A+\sin ^{2} A$ $=1$...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=1$ Solution: We have to prove $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=1$ We know that, $\sin ^{2} A+\cos ^{2} A=1$ So, $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=\sin ^{2} A \frac{\cos ^{2} A}{\sin ^{2} A}+\cos ^{2} A \frac{\sin ^{2} A}{\cos ^{2} A}$ $=\cos ^{2} A+\sin ^{2} A$ $=1$...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\left(1+\tan ^{2} \theta\right)(1-\sin \theta)(1+\sin \theta)=1$ Solution: We have to prove $\left(1+\tan ^{2} \theta\right)(1-\sin \theta)(1+\sin \theta)=1$ We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$ $\sec ^{2} \theta-\tan ^{2} \theta=1$ So, $\left(1+\tan ^{2} \theta\right)(1-\sin \theta)(1+\sin \theta)=\left(1+\tan ^{2} \theta\right)\{(1-\sin \theta)(1+\sin \theta)\}$ $=\left(1+\tan ^{2} \theta\right)\left(1-\sin ^{2} \theta\rig...
Read More →One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
Question: One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent? (i) E: the card drawn is a spade F: the card drawn is an ace (ii) E: the card drawn is black F: the card drawn is a king (iii) E: the card drawn is a king or queen F: the card drawn is a queen or jack Solution: (i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces. $\therefore P(E)=P($ the card drawn is a spade $)=\frac{13}{52}=\frac{1}{4}...
Read More →In a ∆ABC, if
Question: In a ∆ABC, ifc2+a2b2=ac, then the measure of angleBis ____________. Solution: In a ∆ABC, ifc2+a2b2=ac Since $\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 a c}$ $=\frac{a c}{2 a c}$ i.e $\cos B=\frac{1}{2}$ i.e $B=\frac{\pi}{3}$...
Read More →Rationalise the denominator of each of the following.
Question: Rationalise the denominator of each of the following. (i) $\frac{1}{\sqrt{7}}$ (ii) $\frac{\sqrt{5}}{2 \sqrt{3}}$ (iii) $\frac{1}{2+\sqrt{3}}$ (iv) $\frac{1}{\sqrt{5}-2}$ (v) $\frac{1}{5+3 \sqrt{2}}$ (vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$ (vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$ (viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$ (ix) $\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}}$ Solution: (i) $\frac{1}{\sqrt{7}}$ On multiplying the numerator and denominator of the given number by $\sqrt{7}$, we get: $\frac{1}{\sqrt...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta$ Solution: We have to prove $\tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta$ We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$ So, $\tan ^{2} \theta-\sin ^{2} \theta=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-\sin ^{2} \theta$ $=\frac{\sin ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta}{\cos ^{2} \theta}$ $=\frac{\sin ^{2} \theta\left(1-\cos ^{2} \thet...
Read More →In a ∆ABC, if
Question: In a $\Delta A B C$, if $\frac{\sin (A-B)}{\sin (A+B)}=\frac{a^{2}-b^{2}}{k}$, then $k=$ Solution: In a ∆ABC, Given $\frac{\sin (A-B)}{\sin (A+B)}=\frac{\sin A \cos B-\cos A \sin B}{\sin A \cos B+\cos A \sin B}$ Since $\quad \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$ and $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=x$ say i. e $\frac{\sin (A-B)}{\sin (A+B)}=\frac{\left(\frac{a}{x}\right)\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)-\left(\fra...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In any $\Delta \mathrm{ABC}, a(b \cos C-c \cos B)=$ (a) $a^{2}$ (b) $b^{2}-c^{2}$ (c) 0 (d) $b^{2}+c^{2}$ Solution: Using cosine rule, we have $a(b \cos C-c \cos B)$ $=a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)-c a\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)$ $=\frac{a^{2}+b^{2}-c^{2}-c^{2}-a^{2}+b^{2}}{2}$ $=\frac{2 b^{2}-2 c^{2}}{2}$ $=b^{2}-c^{2}$ Hence, the correct answer is option (b)....
Read More →Probability of solving specific problem independently by
Question: Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem. Solution: Probability of solving the problem by $A, P(A)=\frac{1}{2}$ Probability of solving the problem by $B, P(B)=\frac{1}{3}$ Since the problem is solved independently by A and B, $\therefore \mathrm{P}(\mathrm{AB})=\mathr...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In any $\Delta \mathrm{ABC}, a(b \cos C-c \cos B)=$ (a) $a^{2}$ (b) $b^{2}-c^{2}$ (c) 0 (d) $b^{2}+c^{2}$ Solution: Using cosine rule, we have $a(b \cos C-c \cos B)$ $=a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)-c a\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)$ $=\frac{a^{2}+b^{2}-c^{2}-c^{2}-a^{2}+b^{2}}{2}$ $=\frac{2 b^{2}-2 c^{2}}{2}$ $=b^{2}-c^{2}$ Hence, the correct answer is option (b)....
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In any $\triangle \mathrm{ABC}$, the value of $2 a c \sin \left(\frac{A-B+C}{2}\right)$ is (a) $a^{2}+b^{2}-c^{2}$ (b) $c^{2}+a^{2}-b^{2}$ (c) $b^{2}-c^{2}-a^{2}$ (d) $c^{2}-a^{2}-b^{2}$ Solution: In $\Delta \mathrm{ABC}$ $A+B+C=\pi \quad$ (Angle sum property) $\Rightarrow A+C=\pi-B$ $\therefore 2 a c \sin \left(\frac{A-B+C}{2}\right)$ $=2 a c \sin \left(\frac{\pi-2 B}{2}\right)$ $=2 a c \sin \left(\frac{\pi}{2}-B\right)$ $=2 a c \...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In any $\triangle \mathrm{ABC}$, the value of $2 a c \sin \left(\frac{A-B+C}{2}\right)$ is (a) $a^{2}+b^{2}-c^{2}$ (b) $c^{2}+a^{2}-b^{2}$ (c) $b^{2}-c^{2}-a^{2}$ (d) $c^{2}-a^{2}-b^{2}$ Solution: In $\Delta \mathrm{ABC}$ $A+B+C=\pi \quad$ (Angle sum property) $\Rightarrow A+C=\pi-B$ $\therefore 2 a c \sin \left(\frac{A-B+C}{2}\right)$ $=2 a c \sin \left(\frac{\pi-2 B}{2}\right)$ $=2 a c \sin \left(\frac{\pi}{2}-B\right)$ $=2 a c \...
Read More →Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
Question: Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red. Solution: Total number of balls = 18 Number of red balls = 8 Number of black balls = 10 (i) Probability of getting a red ball in the first draw $=\frac{8}{18}=\frac{4}{9}$ The ball is replaced after the first draw. $\therefore$ Probability of getting a red...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In a $\triangle \mathrm{ABC}$, if $(c+a+b)(a+b-c)=a b$, then the measure of angle $C$ is (a) $\frac{\pi}{3}$ (b) $\frac{\pi}{6}$ (c) $\frac{2 \pi}{3}$ (d) $\frac{\pi}{2}$ Solution: Given: $(c+a+b)(a+b-c)=a b$ $\Rightarrow(a+b)^{2}-c^{2}=a b$ $\Rightarrow a^{2}+b^{2}+2 a b-c^{2}=a b$ $\Rightarrow a^{2}+b^{2}-c^{2}=-a b$ $\Rightarrow \frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2}$ $\Rightarrow \cos C=-\frac{1}{2}=\cos \frac{2 \pi}{3} \...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $(\operatorname{cosec} A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1$ Solution: We have to prove $(\operatorname{cosec} A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1$ We know that, $\sin ^{2} A+\cos ^{2} A=1$ So, $(\operatorname{cosec} A-\sin A)(\sec A-\cos A)(\tan A+\cot A)$ $=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)$ $=\left(\frac{1-\sin ^{2} A}{\sin A}\right)...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In a $\triangle \mathrm{ABC}$, if $(c+a+b)(a+b-c)=a b$, then the measure of angle $C$ is (a) $\frac{\pi}{3}$ (b) $\frac{\pi}{6}$ (c) $\frac{2 \pi}{3}$ (d) $\frac{\pi}{2}$ Solution: Given: $(c+a+b)(a+b-c)=a b$ $\Rightarrow(a+b)^{2}-c^{2}=a b$ $\Rightarrow a^{2}+b^{2}+2 a b-c^{2}=a b$ $\Rightarrow a^{2}+b^{2}-c^{2}=-a b$ $\Rightarrow \frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2}$ $\Rightarrow \cos C=-\frac{1}{2}=\cos \frac{2 \pi}{3} \...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\sec A(1-\sin A)(\sec A+\tan A)=1$ Solution: We have to prove $\sec A(1-\sin A)(\sec A+\tan A)=1$ We know that, $\sec ^{2} A-\tan ^{2} A=1$ So, $\sec A(1-\sin A)(\sec A+\tan A)=\{\sec A(1-\sin A)\}(\sec A+\tan A)$ $=(\sec A-\sec A \sin A)(\sec A+\tan A)$ $=\left(\sec A-\frac{1}{\cos A} \sin A\right)(\sec A+\tan A)$ $=\left(\sec A-\frac{\sin A}{\cos A}\right)(\sec A+\tan A)$ $=(\sec A-\tan A)(\sec A+\tan A)$ $=\sec ^{2} A-\tan ^{2} A$ $=1$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In a triangle $\mathrm{ABC}, a=4, b=3, \angle A=60^{\circ}$ then $c$ is a root of the equation (a) $c^{2}-3 c-7=0$ (b) $c^{2}+3 c+7=0$ (c) $c^{2}-3 c+7=0$ (d) $c^{2}+3 c-7=0$ Solution: It is given that $a=4, b=3$ and $\angle A=60^{\circ}$. Using cosine rule, we have $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ $\Rightarrow \cos 60^{\circ}=\frac{9+c^{2}-16}{2 \times 3 \times c}$ $\Rightarrow \frac{1}{2}=\frac{c^{2}-7}{6 c}$ $\Rightarrow...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In a triangle $\mathrm{ABC}, a=4, b=3, \angle A=60^{\circ}$ then $c$ is a root of the equation (a) $c^{2}-3 c-7=0$ (b) $c^{2}+3 c+7=0$ (c) $c^{2}-3 c+7=0$ (d) $c^{2}+3 c-7=0$ Solution: It is given that $a=4, b=3$ and $\angle A=60^{\circ}$. Using cosine rule, we have $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ $\Rightarrow \cos 60^{\circ}=\frac{9+c^{2}-16}{2 \times 3 \times c}$ $\Rightarrow \frac{1}{2}=\frac{c^{2}-7}{6 c}$ $\Rightarrow...
Read More →A die is tossed thrice. Find the probability of getting an odd number at least once.
Question: A die is tossed thrice. Find the probability of getting an odd number at least once. Solution: Probability of getting an odd number in a single throw of a die $=\frac{3}{6}=\frac{1}{2}$ Similarly, probability of getting an even number $=\frac{3}{6}=\frac{1}{2}$ Probability of getting an even number three times $=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}$ Therefore, probability of getting an odd number at least once $=1$ - Probability of getting an odd number in none...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $(\sec \theta+\cos \theta)(\sec \theta-\cos \theta)=\tan ^{2} \theta+\sin ^{2} \theta$ Solution: We have to prove $(\sec \theta+\cos \theta)(\sec \theta-\cos \theta)=\tan ^{2} \theta+\sin ^{2} \theta$ We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$ $\sec ^{2} \theta-\tan ^{2} \theta=1$ $(\sec \theta+\cos \theta)(\sec \theta-\cos \theta)=\sec ^{2} \theta-\cos ^{2} \theta$ $=\left(1+\tan ^{2} \theta\right)-\left(1-\sin ^{2} \theta\right)$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: In any ∆ABC, 2(bccosA+cacosB+abcosC) = (a) $a b c$ (b) $a+b+c$ (c) $a^{2}+b^{2}+c^{2}$ (d) $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$ Solution: Using cosine rule, we have $2(b c \cos A+c a \cos B+a b \cos C)$ $=2 b c\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)+2 c a\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)+2 a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)$ $=b^{2}+c^{2}-a^{2}+c^{2}+a^{2}-b^{2}+a^{2}+b^{2}-c^{2}$ $=a^{2}+b...
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