If sin θ=45, what is the value of cotθ + cosecθ?

Question: If $\sin \theta=\frac{4}{5}$, what is the value of $\cot \theta+\operatorname{cosec} \theta ?$ Solution: Given: $\sin \theta=\frac{4}{5}$ We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$ $\Rightarrow\left(\frac{4}{5}\right)^{2}+\cos ^{2} \theta=1$ $\Rightarrow \frac{16}{25}+\cos ^{2} \theta=1$ $\Rightarrow \cos ^{2} \theta=1-\frac{16}{25}$ $\Rightarrow \cos ^{2} \theta=\frac{9}{25}$ $\Rightarrow \cos \theta=\frac{3}{5}$ We have, $\cot \theta+\operatorname{cosec} \theta=\frac{\cos \t...

Suppose we have four boxes. A, B, C and D containing coloured marbles as given below:

Question: Suppose we have four boxes. A, B, C and D containing coloured marbles as given below: One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C? Solution: Let R be the event of drawing the red marble. Let EA, EB, and ECrespectively denote the events of selecting the box A, B, and C. Total number of marbles = 40 Number of red marbles = 15 $\therefore P(R)=\frac{15}{40}=\fr...

2 + 5 + 8 + 11 + ... + (3n − 1)

Question: $2+5+8+11+\ldots+(3 n-1)=\frac{1}{2} n(3 n+1)$ Solution: LetP(n) be the given statement. Now, $P(n)=2+5+8+\ldots+(3 n-1)=\frac{1}{2} n(3 n+1)$ Step 1: $P(1)=2=\frac{1}{2} \times 1(3+1)$ Hence, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $2+5+8+\ldots+(3 m-1)=\frac{1}{2} m(3 m+1)$ To prove: $P(m+1)$ is true. That is, $2+5+8+\ldots+(3 m+2)=\frac{1}{2}(m+1)(3 m+4)$ $P(m)$ is equal to: $2+5+8+\ldots+(3 m-1)=\frac{1}{2} m(3 m+1)$ Thus, we have : $2+5+8+\ldots+(3 m-1)+(3 m+2)=\frac{1}...

2 + 5 + 8 + 11 + ... + (3n − 1)

Question: $2+5+8+11+\ldots+(3 n-1)=\frac{1}{2} n(3 n+1)$ Solution: LetP(n) be the given statement. Now, $P(n)=2+5+8+\ldots+(3 n-1)=\frac{1}{2} n(3 n+1)$ Step 1: $P(1)=2=\frac{1}{2} \times 1(3+1)$ Hence, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $2+5+8+\ldots+(3 m-1)=\frac{1}{2} m(3 m+1)$ To prove: $P(m+1)$ is true. That is, $2+5+8+\ldots+(3 m+2)=\frac{1}{2}(m+1)(3 m+4)$ $P(m)$ is equal to: $2+5+8+\ldots+(3 m-1)=\frac{1}{2} m(3 m+1)$ Thus, we have : $2+5+8+\ldots+(3 m-1)+(3 m+2)=\frac{1}...

The product of a nonzero rational number with an irrational number is always a/an

Question: The product of a nonzero rational number with an irrational number is always a/an(a) irrational number(b) rational number(c) whole number(d) natural number Solution: Since, the product of a non-zero rational number with an irrational number is always an irrational number.Hence, the correct option is (a)....

If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?

Question: If $x=a \sin \theta$ and $y=b \cos \theta$, what is the value of $b^{2} x^{2}+a^{2} y^{2}$ ? Solution: Given: $x=a \sin \theta$ and $y=b \cos \theta$ So, $b^{2} x^{2}+a^{2} y^{2}=b^{2}(a \sin \theta)^{2}+a^{2}(b \cos \theta)^{2}$ $=a^{2} b^{2} \sin ^{2} \theta+a^{2} b^{2} \cos ^{2} \theta$ $=a^{2} b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$ We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$ Therefore, $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$...

Write the value of cot2 θ−1sin2 θ.

Question: Write the value of $\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}$. Solution: We have, $\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}=\cot ^{2} \theta-\left(\frac{1}{\sin \theta}\right)^{2}$ $=\cot ^{2} \theta-(\operatorname{cosec} \theta)^{2}$ $=\cot ^{2} \theta-\operatorname{cosec}^{2} \theta$ We know that, $\cot ^{2} \theta-\operatorname{cosec}^{2} \theta=-1$ Therefore, $\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}=-1$...

Which of the following is a rational number?

Question: Which of the following is a rational number? (a) $\sqrt{5}$ (b) $0.101001000100001 \ldots$ (c) $\pi$ (d) $0.853853853 \ldots$ Solution: Since,a number whose decimal expansion is terminating or non-terminating recurring is rational number.So,0.853853853... is a rational number.Hence, the correct option is (d)....

In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown.

Question: In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses. Solution: When a die is thrown, then probability of getting a six $=\frac{1}{6}$ then, probability of not getting a six $=1-\frac{1}{6}=\frac{5}{6}$ If the man gets a six in the first throw, then probability of getting a six $=\frac{1}{6}$ If he does not...

Write the value of sin A cos (90° − A) + cos A sin (90° − A).

Question: Write the value of $\sin A \cos \left(90^{\circ}-A\right)+\cos A \sin \left(90^{\circ}-A\right)$. Solution: We have, $\sin A \cos \left(90^{\circ}-A\right)+\cos A \sin \left(90^{\circ}-A\right)=\sin A \sin A+\cos A \cos A$ $=\sin ^{2} A+\cos ^{2} A$ We know that, $\sin ^{2} A+\cos ^{2} A=1$ Therefore, $\sin A \cos \left(90^{\circ}-A\right)+\cos A \sin \left(90^{\circ}-A\right)=1$...

1.2 + 2.2

Question: 1.2 + 2.22+ 3.23+ ... +n.2n= (n 1) 2n+1+2 Solution: LetP(n) be the given statement. Now, $P(n)=1.2+2.2^{2}+3.2^{3}+\ldots+n .2^{n}=(n-1) 2^{n+1}+2$ Step 1: $P(1)=1.2=2=(1-1) 2^{1+1}+2$ Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $1.2+2.2^{2}+\ldots+m .2^{m}=(m-1) 2^{m+1}+2$ To prove: $P(m+1)$ is true. That is, $1.2+2.2^{2}+\ldots+(m+1) 2^{m+1}=m .2^{m+2}+2$ Now, $P(m)=1.2+2.2^{2}+\ldots+m .2^{m}=(m-1) 2^{m+1}+2$ $\Rightarrow 1.2+2.2^{2}+\ldots+m .2^{m}+(m+1) \cdot 2^{m+1}=...

1.2 + 2.2

Question: 1.2 + 2.22+ 3.23+ ... +n.2n= (n 1) 2n+1+2 Solution: LetP(n) be the given statement. Now, $P(n)=1.2+2.2^{2}+3.2^{3}+\ldots+n .2^{n}=(n-1) 2^{n+1}+2$ Step 1: $P(1)=1.2=2=(1-1) 2^{1+1}+2$ Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $1.2+2.2^{2}+\ldots+m .2^{m}=(m-1) 2^{m+1}+2$ To prove: $P(m+1)$ is true. That is, $1.2+2.2^{2}+\ldots+(m+1) 2^{m+1}=m .2^{m+2}+2$ Now, $P(m)=1.2+2.2^{2}+\ldots+m .2^{m}=(m-1) 2^{m+1}+2$ $\Rightarrow 1.2+2.2^{2}+\ldots+m .2^{m}+(m+1) \cdot 2^{m+1}=...

Write the value of cosec2 (90° − θ) − tan2 θ.

Question: Write the value of $\operatorname{cosec}^{2}\left(90^{\circ}-\theta\right)-\tan ^{2} \theta$. Solution: We have, $\operatorname{cosec}^{2}\left(90^{\circ}-\theta\right)-\tan ^{2} \theta=\left\{\operatorname{cosec}\left(90^{\circ}-\theta\right)\right\}^{2}-\tan ^{2} \theta$ $=(\sec \theta)^{2}-\tan ^{2} \theta$ $=\sec ^{2} \theta-\tan ^{2} \theta$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ Therefore, $\operatorname{cosec}^{2}\left(90^{\circ}-\theta\right)-\tan ^{2} \theta=1$...

If cosec θ − cot θ = α, write the value of cosec θ + cot α.

Question: If $\operatorname{cosec} \theta-\cot \theta=\alpha$, write the value of $\operatorname{cosec} \theta+\cot \alpha$ Solution: Given: $\operatorname{cosec} \theta-\cot \theta=\alpha$ We know that, $\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1$ Therefore, $\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1$ $\Rightarrow(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)=1$ $\Rightarrow(\operatorname{cosec} \theta+\cot \theta) \alpha=1$ $\Rightarrow(\ope...

Question: $\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.5}+\ldots+\frac{1}{(4 n-1)(4 n+3)}=\frac{n}{3(4 n+3)}$ Solution: LetP(n) be the given statement. Now, $P(n)=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\ldots+\frac{1}{(4 n-1)(4 n+3)}=\frac{n}{3(4 n+3)}$ Step 1: $P(1)=\frac{1}{3.7}=\frac{1}{21}=\frac{1}{3(4+3)}$ Hence, $P(1)$ is true. Step 2: Let $P(m)$ is true. Then, $\frac{1}{3.7}+\frac{1}{7.11}+\ldots+\frac{1}{(4 m-1)(4 m+3)}=\frac{m}{3(4 m+3)}$ To prove: $P(m+1)$ is true. That is, $\frac{1...

If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.

Question: If $\sec \theta+\tan \theta=x$, write the value of $\sec \theta-\tan \theta$ in terms of $x .$ Solution: Given: $\sec \theta+\tan \theta=x$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ Therefore, $\sec ^{2} \theta-\tan ^{2} \theta=1$ $\Rightarrow \quad(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1$ $\Rightarrow \quad x(\sec \theta-\tan \theta)=1$ $\Rightarrow \quad(\sec \theta-\tan \theta)=\frac{1}{x}$ Hence, $\sec \theta-\tan \theta=\frac{1}{x}$...

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Question: How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%? Solution: Let the man toss the coinntimes. Thentosses arenBernoulli trials. Probability $(p)$ of getting a head at the toss of a coin is $\frac{1}{2}$. $\therefore p=\frac{1}{2} \Rightarrow q=\frac{1}{2}$ $\therefore \mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} p^{n-x} q^{x}={ }^{n} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{n-x}\left(\frac{1}{2}\right)^{x}={ }^{n} \mathrm{C...

A rational number lying between

Question: A rational number lying between $\sqrt{2}$ and $\sqrt{3}$ is (a) $\frac{(\sqrt{2}+\sqrt{3})}{2}$ (b) $\sqrt{6}$ (c) $1.6$ (d) $1.9$ Solution: Since, $\frac{(\sqrt{2}+\sqrt{3})}{2}$ and $\sqrt{6}$ are irrational numbers, And, $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$ So, the rational number lying between $\sqrt{2}$ and $\sqrt{3}$ is $1.6$. Hence, the correct option is (c)....

What is the value of sin2 θ+11+tan2 θ?

Question: What is the value of $\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta} ?$ Solution: We have, $\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}=\sin ^{2} \theta+\frac{1}{\sec ^{2} \theta}$ $=\sin ^{2} \theta+\left(\frac{1}{\sec \theta}\right)^{2}$ $=\sin ^{2} \theta+\cos ^{2} \theta$ $=1$...

What is the value of (1 + cot2 θ) sin2 θ?

Question: What is the value of $\left(1+\cot ^{2} \theta\right) \sin ^{2} \theta ?$ Solution: We have, $\left(1+\cot ^{2} \theta\right) \sin ^{2} \theta=\operatorname{cosec}^{2} \theta \times \sin ^{2} \theta$ $=\left(\frac{1}{\sin \theta}\right)^{2} \times \sin ^{2} \theta$ $=\frac{1}{\sin ^{2} \theta} \times \sin ^{2} \theta$ $=1$...

What is the value of (1 − cos2 θ) cosec2 θ?

Question: What is the value of $\left(1-\cos ^{2} \theta\right) \operatorname{cosec}^{2} \theta ?$ Solution: We have, $\left(1-\cos ^{2} \theta\right) \operatorname{cosec}^{2} \theta=\sin ^{2} \theta \times \operatorname{cosec}^{2} \theta$ $=\sin ^{2} \theta \times\left(\frac{1}{\sin \theta}\right)^{2}$ $=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$ $=1$...

Which of the following is a true statment?

Question: Which of the following is a true statment? (a) $\pi$ and $\frac{22}{7}$ are both rationals (b) $\pi$ and $\frac{22}{7}$ are both irrationals (c) $\pi$ is rational and $\frac{22}{7}$ is irrational (d) $\pi$ is irrational and $\frac{22}{7}$ is rational Solution: (d) $\pi$ is irrational and $\frac{22}{7}$ is rational. Because the value of $\pi$ is neither repeating nor terminating, it is an irrational number. $\frac{22}{7}$, on the other hand, is of the form $\frac{p}{q}$, so it is a rati...

Define an identity.

Question: Define an identity. Solution: An identity is an equation which is true for all values of the variable (s). For example, $(x+3)^{2}=x^{2}+6 x+9$ Any number of variables may involve in an identity. An example of an identity containing two variables is $(x+y)^{2}=x^{2}+2 x y+y^{2}$ The above are all about algebraic identities. Now, we define the trigonometric identities. An equation involving trigonometric ratios of an angle(say) is said to be a trigonometric identity if it is satisfied f...

Which of the following is a true statment?

Question: Which of the following is a true statment?(a) The sum of two irrational numbers is an irrational number(b) The product of two irrational numbers is an irrational number(c) Every real number is always rational(d) Every real number is either rational or irrational Solution: (d) Every real number is either rational or irrational.Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational....

Question: $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}$ Solution: LetP(n) be the given statement. Now, $P(n)=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}$ Step 1; $P(1)=\frac{1}{3.5}=\frac{1}{15}=\frac{1}{3(2+3)}$ Hence, $P(1)$ is true. Step 2: Let $P(m)$ be true. Then, $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 m+1)(2 m+3)}=\frac{m}{3(2 m+3)}$ To prove : $P(m+1)$ is true. That is,...