- A. $\cos^2 \theta – \sin^2 \theta = 1$
- B. $\operatorname{cosec}^2 \theta – \sec^2 \theta = 1$
- C. $\sec^2 \theta – \tan^2 \theta = 1$
- D. $\cot^2 \theta – \tan^2 \theta = 1$
The trigonometric identity $\sec^2 \theta - \tan^2 \theta = 1$ is true for all values of $\theta$ where the functions are defined ($0^{\circ} \le \theta < 90^{\circ}$). Option (a) is incorrect because $\cos^2 \theta + \sin^2 \theta = 1$. Option (b) and (d) are not standard identities.
- A. 3
- B. – 3
- C. 4
- D. – 4
For three terms $a, b, c$ to be in A.P., $2b = a + c$. Substituting the given terms: $2(4k - 6) = (k + 2) + (3k - 2) \implies 8k - 12 = 4k \implies 4k = 12 \implies k = 3$.
- A. 12 cm
- B. 20 cm
- C. 6 cm
- D. 14 cm
By Thales' Theorem (Basic Proportionality Theorem), since $PQ \parallel BC$, we have $\frac{AP}{PB} = \frac{AQ}{QC}$. Substituting values: $\frac{4}{6} = \frac{8}{QC} \implies 4 \cdot QC = 48 \implies QC = 12$. The length $AC = AQ + QC = 8 + 12 = 20$ cm.
- A. 1:2
- B. 2:1
- C. 1:1
- D. 1:3
The least composite number is 4 and the least prime number is 2. HCF(4, 2) = 2. LCM(4, 2) = 4. The ratio of HCF to LCM is $2 : 4 = 1 : 2$.
- A. $\frac{1}{13}$
- B. $\frac{9}{13}$
- C. $\frac{4}{13}$
- D. $\frac{12}{13}$
There are 4 aces in a pack of 52 cards. The number of cards that are not aces is $52 - 4 = 48$. Probability $P(\text{not an ace}) = \frac{48}{52} = \frac{12}{13}$.
- A. 3.6 cm
- B. 2.5 cm
- C. 10 cm
- D. 3.2 cm
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Log in to view solution →- A. 2, –5
- B. –2, 5
- C. 2, 5
- D. –2, –5
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Log in to view solution →- A. 60$^{\circ}$
- B. 45$^{\circ}$
- C. 30$^{\circ}$
- D. 90$^{\circ}$
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Log in to view solution →- A. 6
- B. – 6
- C. 8
- D. 10
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Log in to view solution →- A. $\frac{1}{16} \pi d^2$
- B. $\frac{1}{4} \pi d^2$
- C. $\frac{1}{8} \pi d^2$
- D. $\frac{1}{2} \pi d^2$
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Log in to view solution →- A. 15
- B. 25
- C. 30
- D. 35
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Log in to view solution →- A. 40 cm
- B. 9 cm
- C. 41 cm
- D. 50 cm
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Log in to view solution →- A. 50$^{\circ}$
- B. 40$^{\circ}$
- C. 100$^{\circ}$
- D. 130$^{\circ}$
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Log in to view solution →- A. 15
- B. 25
- C. 30
- D. 35
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Log in to view solution →- A. 18
- B. 15
- C. 10
- D. 20
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Log in to view solution →- A. 2
- B. 1
- C. –1
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Log in to view solution →- A. $\frac{7}{3}$
- B. $-\frac{7}{3}$
- C. $\frac{3}{7}$
- D. $-\frac{3}{7}$
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Log in to view solution →- A. intersecting
- B. parallel
- C. coincident
- D. either intersecting or parallel
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Log in to view solution →- A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- B. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
- C. Assertion (A) is true but Reason (R) is false.
- D. Assertion (A) is false but Reason (R) is true.
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Log in to view solution →- A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- B. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
- C. Assertion (A) is true but Reason (R) is false.
- D. Assertion (A) is false but Reason (R) is true.
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Log in to view solution →Jagdish has a field which is in the shape of a right angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O.
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Log in to view solution →Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking. After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats.
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Log in to view solution →Two schools ‘P’ and ‘Q’ decided to award prizes to their students for two games of Hockey ` x per student and Cricket ` y per student. School ‘P’ decided to award a total of ` 9,500 for the two games to 5 and 4 students respectively; while school ‘Q’ decided to award ` 7,370 for the two games to 4 and 3 students respectively.
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