The value of 1+cos θ1−cos θ−−−−−−√ is

Question: The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is (a) $\cot \theta-\operatorname{cosec} \theta$ (b) $\operatorname{cosec} \theta+\cot \theta$ (c) $\operatorname{cosec}^{2} \theta+\cot ^{2} \theta$ (d) $(\cot \theta+\operatorname{cosec} \theta)^{2}$ Solution: The given expression is $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$. Multiplying both the numerator and denominator under the root by $(1+\cos \theta)$, we have $\sqrt{\frac{(1+\cos \theta)(1+\cos \theta)}{(1+\cos \thet...

Solve the following

Question: 32n+7 is divisible by 8 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 3^{2 n}+7$ is divisible by 8 for all $n \in N$ Step 1: $P(1)=3^{2}+7=9+7=16$ It is divisible by 8 . Step 2: Let $P(m)$ be true. Then, $3^{2 m}+7$ is divisible by 8 . T hus, $3^{2 m}+7=8 \lambda$ for some $\lambda \in N$. ....(1) We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=3^{2 m+2}+7$ $=3^{2 m} \cdot 9+7$ $=(8 \lambda-7) \cdot 9+7$ From (1) $=72 \lambda-63+7$ $=72 \l...

Solve the following

Question: 32n+7 is divisible by 8 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 3^{2 n}+7$ is divisible by 8 for all $n \in N$ Step 1: $P(1)=3^{2}+7=9+7=16$ It is divisible by 8 . Step 2: Let $P(m)$ be true. Then, $3^{2 m}+7$ is divisible by 8 . T hus, $3^{2 m}+7=8 \lambda$ for some $\lambda \in N$. ....(1) We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=3^{2 m+2}+7$ $=3^{2 m} \cdot 9+7$ $=(8 \lambda-7) \cdot 9+7$ From (1) $=72 \lambda-63+7$ $=72 \l...

1+sin θ1−sin θ−−−−−√ is equal to

Question: $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$ is equal to (a) $\sec \theta+\tan \theta$ (b) $\sec \theta-\tan \theta$ (c) $\sec ^{2} \theta+\tan ^{2} \theta$ (d) $\sec ^{2} \theta-\tan ^{2} \theta$ Solution: The given expression is $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$. Multiplying both the numerator and denominator under the root by $(1+\sin \theta)$, we have $\sqrt{\frac{(1+\sin \theta)(1+\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}}$ $\sqrt{(1+\sin \theta)(1-\sin \theta)}$ $=...

The value of

Question: The value of $2 . \overline{45}+0 . \overline{36}$ is (a) $\frac{67}{33}$ (b) $\frac{24}{11}$ (c) $\frac{31}{11}$ (d) $\frac{167}{110}$ Solution: Let $x=2 . \overline{45}=2.4545 \ldots \quad \ldots(1)$ Multiplying both sides by 100, we get $100 x=245 . \overline{45} \quad \ldots \ldots(2)$ Subtracting (1) from (2), we get $100 x-x=245 . \overline{45}-2 . \overline{45}$ $\Rightarrow 99 x=245-2=243$ $\Rightarrow x=\frac{243}{99}$ $\therefore 2 . \overline{45}=\frac{243}{99}$ Let $y=0 . \...

if secθ+tanθ=x, then tanθ=

Question: If $\sec \theta+\tan \theta=x$, then $\tan \theta=$ (a) $\frac{x^{2}+1}{x}$ (b) $\frac{x^{2}-1}{x}$ (c) $\frac{x^{2}+1}{2 x}$ (d) $\frac{x^{2}-1}{2 x}$ Solution: Given: $\sec \theta+\tan \theta=x$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ $\Rightarrow(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1$ $\Rightarrow x(\sec \theta-\tan \theta)=1$ $\Rightarrow \sec \theta-\tan \theta=\frac{1}{x}$ Now, $\sec \theta+\tan \theta=x$ $\sec \theta-\tan \theta=\frac{1}{x}$ Subtracting...

If sec θ + tan θ = x, then sec θ =

Question: If $\sec \theta+\tan \theta=x$, then $\sec \theta=$ (a) $\frac{x^{2}+1}{x}$ (b) $\frac{x^{2}+1}{2 x}$ (c) $\frac{x^{2}-1}{2 x}$ (d) $\frac{x^{2}-1}{x}$ Solution: Given: $\sec \theta+\tan \theta=x$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ $\Rightarrow(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1$ $\Rightarrow x(\sec \theta-\tan \theta)=1$ $\Rightarrow \sec \theta-\tan \theta=\frac{1}{x}$ Now, $\sec \theta+\tan \theta=x$ $\sec \theta-\tan \theta=\frac{1}{x}$ Adding the ...

Question: 52n1 is divisible by 24 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 5^{2 n}-1$ is divisible by 24 for all $n \in N$. Step 1: $P(1)=5^{2}-1=25-1=24$ It is divisible by 24 . Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $5^{2 m}-1$ is divisible by 24 . Now, let $5^{2 m}-1=24 \lambda$, where $\lambda \in N$. We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=5^{2 m+2}-1$ $=5^{2 m} 5^{2}-1$ $=25(24 \lambda+1)-1$ $=600 \lambda+24$ $=2...

Question: 52n1 is divisible by 24 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 5^{2 n}-1$ is divisible by 24 for all $n \in N$. Step 1: $P(1)=5^{2}-1=25-1=24$ It is divisible by 24 . Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $5^{2 m}-1$ is divisible by 24 . Now, let $5^{2 m}-1=24 \lambda$, where $\lambda \in N$. We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=5^{2 m+2}-1$ $=5^{2 m} 5^{2}-1$ $=25(24 \lambda+1)-1$ $=600 \lambda+24$ $=2...

If cosec θ = 2x and cot θ=2x, find the value of 2(x2−1x2).

Question: If $\operatorname{cosec} \theta=2 x$ and $\cot \theta=\frac{2}{x}$, find the value of $2\left(x^{2}-\frac{1}{x^{2}}\right) .$ Solution: Given: $\operatorname{cosec} \theta=2 x, \cot \theta=\frac{2}{x}$ We know that, $\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1$ $\Rightarrow(2 x)^{2}-\left(\frac{2}{x}\right)^{2}=1$ $\Rightarrow \quad 4 x^{2}-\frac{4}{x^{2}}=1$ $\Rightarrow 4\left(x^{2}-\frac{1}{x^{2}}\right)=1$ $\Rightarrow \quad 2 \times 2 \times\left(x^{2}-\frac{1}{x^{2}}\right...

a + (a + d) + (a + 2d) + ... (a + (n − 1) d)

Question: $a+(a+d)+(a+2 d)+\ldots(a+(n-1) d)=\frac{n}{2}[2 a+(n-1) d]$ Solution: LetP(n) be the given statement. Now, $P(n): a+(a+d)+(a+2 d)+\ldots+(a+(n-1) d)=\frac{n}{2}[2 a+(n-1) d]$ Step 1: $P(1)=a=\frac{1}{2}(2 a+(1-1) d)$ Hence, $P(1)$ is true. Step 2: Suppose $P(m)$ is true. Then, $a+(a+d)+\ldots+(a+(m-1) d)=\frac{m}{2}[2 a+(m-1) d]$ We have to show that $P(m+1)$ is true whenever $P(m)$ is true. That is, $a+(a+d)+\ldots+(a+m d)=\frac{(m+1)}{2}[2 a+m d]$ $W e$ know that $P(m)$ is true. Thu...

If 5x = sec θ and 5x=tan θ, find the value of 5(x2−1x2).

Question: If $5 x=\sec \theta$ and $\frac{5}{x}=\tan \theta$, find the value of $5\left(x^{2}-\frac{1}{x^{2}}\right)$ Solution: Given: $5 x=\sec \theta, \frac{5}{x}=\tan \theta$ $\Rightarrow \sec \theta=5 x, \tan \theta=\frac{5}{x}$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ $\Rightarrow(5 x)^{2}-\left(\frac{5}{x}\right)^{2}=1$ $\Rightarrow 25 x^{2}-\frac{25}{x^{2}}=1$ $\Rightarrow 5 \times 5 \times\left(x^{2}-\frac{1}{x^{2}}\right)=1$ $\Rightarrow 5\left(x^{2}-\frac{1}{x^{2}}\right)=\f...

a + (a + d) + (a + 2d) + ... (a + (n − 1) d)

Question: $a+(a+d)+(a+2 d)+\ldots(a+(n-1) d)=\frac{n}{2}[2 a+(n-1) d]$ Solution: LetP(n) be the given statement. Now, $P(n): a+(a+d)+(a+2 d)+\ldots+(a+(n-1) d)=\frac{n}{2}[2 a+(n-1) d]$ Step 1: $P(1)=a=\frac{1}{2}(2 a+(1-1) d)$ Hence, $P(1)$ is true. Step 2: Suppose $P(m)$ is true. Then, $a+(a+d)+\ldots+(a+(m-1) d)=\frac{m}{2}[2 a+(m-1) d]$ We have to show that $P(m+1)$ is true whenever $P(m)$ is true. That is, $a+(a+d)+\ldots+(a+m d)=\frac{(m+1)}{2}[2 a+m d]$ $W e$ know that $P(m)$ is true. Thu...

The sum of

Question: The sum of $0 . \overline{3}$ and $0 . \overline{4}$ is (a) $\frac{7}{10}$ (b) $\frac{7}{9}$ (c) $\frac{7}{11}$ (d) $\frac{7}{99}$ Solution: Let $x=0 . \overline{3}=0.333 \ldots \quad \ldots(1)$ Multiplying both sides by 10, we get $10 x=3 . \overline{3} \quad \ldots . .(2)$ Subtracting (1) from (2), we get $10 x-x=3 . \overline{3}-0 . \overline{3}$ $\Rightarrow 9 x=3$ $\Rightarrow x=\frac{3}{9}$ $\therefore 0 . \overline{3}=\frac{3}{9}$ Let $y=0 . \overline{4}=0.444 \ldots \quad \ldot...

If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.

Question: If $\sin ^{2} \theta \cos ^{2} \theta\left(1+\tan ^{2} \theta\right)\left(1+\cot ^{2} \theta\right)=\lambda$, then find the value of $\lambda .$ Solution: Given: $\sin ^{2} \theta \cos ^{2} \theta\left(1+\tan ^{2} \theta\right)\left(1+\cot ^{2} \theta\right)=\lambda$ $\Rightarrow \quad \sin ^{2} \theta \cos ^{2} \theta \sec ^{2} \theta \operatorname{cosec}^{2} \theta=\lambda$ $\Rightarrow \quad\left(\sin ^{2} \theta \operatorname{cosec}^{2} \theta\right) \times\left(\cos ^{2} \theta \s...

If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.

Question: If $\operatorname{cosec}^{2} \theta(1+\cos \theta)(1-\cos \theta)=\lambda$, then find the value of $\lambda .$ Solution: Given: $\operatorname{cosec}^{2} \theta(1+\cos \theta)(1-\cos \theta)=\lambda$ $\Rightarrow \operatorname{cosec}^{2} \theta\{(1+\cos \theta)(1-\cos \theta)\}=\lambda$ $\Rightarrow \quad \operatorname{cosec}^{2} \theta\left(1-\cos ^{2} \theta\right)=\lambda$ $\Rightarrow \quad \operatorname{cosec}^{2} \theta \sin ^{2} \theta=\lambda$ $\Rightarrow \quad \frac{1}{\sin ^...

a + ar + ar

Question: $a+a r+a r^{2}+\ldots+a r^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right), r \neq 1$ Solution: LetP(n) be the given statement. Now, Step 1; $P(n)=a+a r+a r^{2}+\ldots+a r^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right), r \neq 1$ $P(1)=a=a\left(\frac{r^{1}-1}{r-1}\right)$ Hence, $P(1)$ is true. Step 2 : Suppose $P(m)$ is true. Then, $a+a r+a r^{2}+\ldots+a r^{m-1}=a\left(\frac{r^{m}-1}{r-1}\right), r \neq 1$ To show: $P(m+1)$ is true whenever $P(m)$ is true. That is, $a+a r+a r^{2}+\ldots+a r^{m}=a\le...

a + ar + ar

Question: $a+a r+a r^{2}+\ldots+a r^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right), r \neq 1$ Solution: LetP(n) be the given statement. Now, Step 1; $P(n)=a+a r+a r^{2}+\ldots+a r^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right), r \neq 1$ $P(1)=a=a\left(\frac{r^{1}-1}{r-1}\right)$ Hence, $P(1)$ is true. Step 2 : Suppose $P(m)$ is true. Then, $a+a r+a r^{2}+\ldots+a r^{m-1}=a\left(\frac{r^{m}-1}{r-1}\right), r \neq 1$ To show: $P(m+1)$ is true whenever $P(m)$ is true. That is, $a+a r+a r^{2}+\ldots+a r^{m}=a\le...

If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.

Question: If $\sec ^{2} \theta(1+\sin \theta)(1-\sin \theta)=k$, then find the value of $k$. Solution: Given: $\sec ^{2} \theta(1+\sin \theta)(1-\sin \theta)=k$ $\Rightarrow \sec ^{2} \theta\{(1+\sin \theta)(1-\sin \theta)\}=k$ $\Rightarrow \quad \sec ^{2} \theta\left(1-\sin ^{2} \theta\right)=k$ $\Rightarrow \quad \sec ^{2} \theta \cos ^{2} \theta=k$ $\Rightarrow \quad \frac{1}{\cos ^{2} \theta} \times \cos ^{2} \theta=k$ $\Rightarrow \quad 1=k$ $\Rightarrow \quad k=1$ Hence, the value ofkis 1....

If A and B are any two events such that

Question: If $A$ and $B$ are any two events such that $P(A)+P(B)-P(A$ and $B)=P(A)$, then (A) $P(B \mid A)=1$ (B) $P(A \mid B)=1$ (C) $P(B \mid A)=0$ (D) $P(A \mid B)=0$ Solution: $P(A)+P(B)-P(A$ and $B)=P(A)$ $\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A})$ $\Rightarrow \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$ $\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{B})$ $\therefore P(...

If cos θ=34, then find the value of 9tan2 θ + 9.

Question: If $\cos \theta=\frac{3}{4}$, then find the value of $9 \tan ^{2} \theta+9$ Solution: Given: $\cos \theta=\frac{3}{4}$ $\Rightarrow \frac{1}{\cos \theta}=\frac{4}{3}$ $\Rightarrow \sec \theta=\frac{4}{3}$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ $\Rightarrow\left(\frac{4}{3}\right)^{2}-\tan ^{2} \theta=1$ $\Rightarrow \tan ^{2} \theta=\frac{16}{9}-1$ $\Rightarrow \tan ^{2} \theta=\frac{7}{9}$ Therefore, $9 \tan ^{2} \theta+9=9 \times \frac{7}{9}+9$ $=7+9$ $=16$...

If P (A|B) > P (A), then which of the following is correct:

Question: If $P(A \mid B)P(A)$, then which of the following is correct: (A) $P(B \mid A)P(B)$ (B) $P(A \cap B)P(A) \cdot P(B)$ (C) $P(B \mid A)P(B)$ (D) $P(B \mid A)=P(B)$ Solution: $P(A \mid B)P(A)$ $\Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\mathrm{P}(\mathrm{A})$ $\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$ $\Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\mathrm{...

If sin θ=13, then find the value of 2cot2 θ + 2.

Question: If $\sin \theta=\frac{1}{3}$, then find the value of $2 \cot ^{2} \theta+2$ Solution: Given: $\sin \theta=\frac{1}{3}$ $\Rightarrow \frac{1}{\sin \theta}=3$ $\Rightarrow \operatorname{cosec} \theta=3$ We know that, $\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1$ $\Rightarrow(3)^{2}-\cot ^{2} \theta=1$ $\Rightarrow \cot ^{2} \theta=9-1$ $\Rightarrow \cot ^{2} \theta=8$ Therefore, $2 \cot ^{2} \theta+2=2 \times 8+2$ $=16+2$ $=18$...

Question: $1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\frac{1}{3} n\left(4 n^{2}-1\right)$ Solution: LetP(n) be the given statement. Now, $P(n)=1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\frac{1}{3} n\left(4 n^{2}-1\right)$ Step 1: $P(1)=1^{2}=1=\frac{1}{3} \times 1 \times(4-1)$ Hence, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $1^{2}+3^{2}+\ldots+(2 m-1)^{2}=\frac{1}{3} m\left(4 m^{2}-1\right)$ To prove : $P(m+1)$ is true whenever $P(m)$ is true. That $i s$ $1^{2}+3^{2}=\ldots+(2 m+1)^{2}=\frac{1}{...

An irrational number between

Question: An irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is (a) $\frac{1}{2}\left(\frac{1}{7}+\frac{2}{7}\right)$ (b) $\left(\frac{1}{7} \times \frac{2}{7}\right)$ (c) $\sqrt{\frac{1}{7} \times \frac{2}{7}}$ (d) none of these Solution: (c) $\sqrt{\frac{1}{7} \times \frac{2}{7}}$ An irrational number between $a$ and $b$ is given as $\sqrt{a b}$....