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Q) Evaluate 2sec2 θ + 3 cosec2 θ – 2 sin θ cos θ if θ = 450. Ans: Since θ = 450, sec 450 = √2, cosec 450 = √2, sin 450 = ; cos 450 = 2sec2 θ + 3 cosec2 θ – 2 sin θ cos θ = 2 (√2)2 + 3 (√2)2 […]
Evaluate 2sec^2 θ + 3 cosec^2 θ – 2 sin θ cos θ if θ = 45. Read More »
Q) In the given figure, ∠ADC = ∠BCA, prove that Δ ACB ~ Δ ADC. Hence find if AC = 8 cm and AD = 3 cm. Ans: Ans: In Δ ACB and Δ ADC, ∠ADC = ∠BCA  (given) ∠A = ∠A (common angle)   Δ ACB ~ Δ
Q) From a solid cylinder of height 20 cm and diameter 12 cm, a conical cavity of height 8 cm and radius 6 cm is hallowed out. Find the total surface area of the remaining solid. Ans:Â Let’s draw the diagram to capture the shape: Total surface area of the cylinder: = curved surface area of
Q) Prove that √3 is an irrational number. (Q30 – 30/1/3 – CBSE 2026 Question Paper) Ans: Let us assume that √3 is a rational number Let √3 = ; where q ≠0 and let p, q are co-primes.  3q2 = p2………………. (i) It means p2 is divisible by 3 p is divisible by 3
Q) Rohan repays his total loan of Rs.1,18,000 by paying every month starting with the first installment of Rs. 1,000. If he increases the installment by Rs. 100 every month, what amount will be paid by him in the 30th installment? What amount of loan has he paid after 30th installment? Ans: We can see
Q) The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term. Ans: Given that a = 15 and Sum of 15 terms of an A.P. S15 = 750 We know that sum of n terms of an A.P. Sn = (2a + (n-1) d)
Q)Â Find the discriminant of the quadratic equation 4×2Â – 5 = 0 and hence comment on the nature of the equation. Ans: Given equation is 4 x2 – 5 = 0 or we can say that, Find the discriminant of 4 x2 + 0 x – 5 = 0 Let’s compare it with standard quadratic
Q) If θ is an acute angle and sin θ = cos θ, find the value of tan2 θ + cot2 θ – 2. Ans: tan2 θ + cot2 θ – 2. = + – 2 By sin θ = cos θ (given), = + – 2 = 1 + 1 – 2 = 0
If θ is an acute angle and sin θ = cos θ, find the value of tan^2 θ + cot^2 θ – 2. Read More »
Q) Evaluate (5/cot2 30°) + (1/ sin2 60°) – cot2 45° + 2 sin2 90° Ans: Givent that, – cot2 45 + 2 sin2 90 = + – (1)2 + 2 (1)2 = + – 1 + 2 = 3 – 1 + 2 = 4
Evaluate (5/cot^2 30) + (1/ sin^2 60) – cot^2 45 + 2 sin^2 90 Read More »
Q) 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